Hyper Packer (reverse)

Hyper Packer was a reverse challenge from FCSC 2022, of medium difficulty, and for which I got first blood.

We were asked to connect to a remote server that sends us multiple binaries, to unpack these and send back the secret they display.

Preliminary analysis

First, we connect to the remote server once and solve the proof-of-work using hashcash, in order to retrieve a sample binary. Surprise, it's a Windows executable.

╭─face@0xff ~/ctf/fcsc/reverse/hyperpacker                                                       
╰─$ file sample.exe                                                                              
sample.exe: PE32+ executable (console) x86-64, for MS Windows

If we try to execute it, nothing really seems to happen. It just runs indefinitely.

Let's open up IDA. Although the executable is ~300KB, there are very few functions:

The main function looks like this:

It first calls sub_140097203, and if it succeeded, calls sub_1400973F0 next. We notice these two sub-functions have an address as argument: 0x14004E000, which is the start of the .data section. If we look at what's in there, all we can see is lots of high entropy data, which accounts for the vast majority of the file's size.

Let's check out the second function first (sub_1400973F0) because it will be easier to understand.

_BOOL8 __fastcall sub_1400973F0(__int64 data)
{
  _BOOL8 result; // rax
  __int64 v2; // [rsp+128h] [rbp-18h]
  __int64 v3; // [rsp+130h] [rbp-10h]

  result = 0;
  if ( CheckMZ((_QWORD *)data) )
  {
    v2 = CheckPE(data);
    if ( v2 )
    {
      v3 = sub_14009773A(v2, data, 0x48A7Ci64);
      if ( v3 )
      {
        if ( sub_1400974A6(v3) && sub_140097647(v2, data, 0x48A7Ci64) )
          result = 1;
      }
    }
  }
  return result;
}

The functions CheckMZ and CheckPE simply check that the data buffer starts with MZ and also contains a PE header. Therefore, these are checks for a valid executable file. We understand another executable shall be unpacked, or decrypted, to this buffer.

We also guess the next functions take care of loading this new executable and running it, hence we will rename sub_1400973F0 as CheckAndRun.

Let's now dive into the sub_140097203 function, which we suppose is what unpacks the executable.

Reversing the unpacking algorithm

Let's go through the Unpack function.

First, a 16-byte buffer (v6) is set to all null bytes. This could very be a key for a block cipher like AES. We will rename it key for the future.

char *v1; // rdi
__int64 v2; // rcx
char v6[16]; // [rsp+118h] [rbp-18h] BYREF

v1 = v6;
v2 = 16i64;
do
{
  *v1++ = 0;
  --v2;
}
while ( v2 );

Next, 0x48A80 bytes of memory are allocated and the data buffer is copied to this newly allocated space.

v3 = VirtualAlloc(0i64, 0x48A80ui64, 0x3000u, 4u);
if ( v3 )
{
  lpAddress = v3;
  qmemcpy(v3, data, 0x48A80ui64);
LABEL_5:
  if ( sub_1400971C8((__int64)data, (__int64)v6) )
  {
    sub_140097132(lpAddress, data, v6, 18600i64);
    LODWORD(v4) = VirtualFree(lpAddress, 0i64, 0x8000u);
    if ( v4 )
      return 1i64;
  }
  else
  {
    qmemcpy(data, lpAddress, 0x10ui64);
    while ( sub_1400973B3(v6) )
    {
      if ( !sub_140097358(v6, 0x140096A80i64, 16i64) )
        goto LABEL_5;
    }
  }
}
return 0i64;

The remaining part makes more sense if we write it with a while loop:

while (1) {
  if (sub_1400971C8(data, key)) {
    sub_140097132(lpAddress, data, key, 0x48A8);
    VirtualFree(lpAddress, 0, 0x8000);
    break;
  } else {
    qmemcpy(data, lpAddress, 0x10ui64);
    while (sub_1400973B3(key)) {
      if (!sub_140097358(key, 0x140096A80, 16))
        break;
    }
  }
}

The sub_1400971C8 function takes the data and key buffers as input. It is rather straightfoward:

_BOOL8 __fastcall sub_1400971C8(__int64 data, __int64 key)
{
  _QWORD outbuf[2]; // [rsp+10h] [rbp-10h] BYREF

  sub_140097132(data, outbuf, key, 1i64);
  return CheckMZ(outbuf);
}

It calls sub_140097132, which is the same function that is called in Unpack right after this the call to this function, although with different arguments: 0x48A8 instead of 1 here.

We notice 0x48A8 is 0x48A80 divided by 16, in other words 0x48A8 is the number of 16-byte blocks in the data buffer, which supports the idea of AES-like symmetrical encryption.

Then, it checks whether outbuf starts with "MZ". This surely means sub_140097132 decrypts a first block of data, and a basic heuristic is used to determine whether the decryption was successful. We will rename this function to GoodFirstBlock, and the sub_140097132 function to DecryptData.

If it passes the check for the first block, then it proceeds to decrypt the entire data buffer. If it does not, then something happens with the key:

while (sub_1400973B3(key)) {
  if (!sub_140097358(key, 0x140096A80, 16))
    break;
}

The first function, sub_1400973B3, is easy to understand:

__int64 __fastcall sub_1400973B3(_BYTE *key)
{
  _BYTE *v1; // rax

  v1 = key;
  do
  {
    if ( ++*v1 != 0xFF )
      return 1i64;
    *v1++ = 0;
  }
  while ( v1 != key + 16 );
  return 0i64;
}

It increments the key buffer, viewed as a 16-byte integer in little endian. We will rename it to Increment.

The second function, sub_140097358, takes the key and an address (0x140096A80) as input:

.data:0000000140096A80 unk_140096A80   db    0                 ; DATA XREF: DecryptData+54↓o
.data:0000000140096A81                 db  98h ; ˜
.data:0000000140096A82                 db    0
.data:0000000140096A83                 db    0
.data:0000000140096A84                 db 0C8h ; È
.data:0000000140096A85                 db    0
.data:0000000140096A86                 db    0
.data:0000000140096A87                 db    0
.data:0000000140096A88                 db    0
.data:0000000140096A89                 db    0
.data:0000000140096A8A                 db    0
.data:0000000140096A8B                 db    0
.data:0000000140096A8C                 db    0
.data:0000000140096A8D                 db    0
.data:0000000140096A8E                 db    0
.data:0000000140096A8F                 db    0

What it does is ensure that all the bits set to 1 in the key lie at positions defined by a certain mask (unk_140096A80). For instance (with a reduced number of bits), with the mask 01100010, the key 01000010 passes the check, but the key 01100011 does not. We will rename this function to CheckBitMask.

Therefore, the following loop increments the key until it finds a key that satisfies the mask.

while (Increment(key)) {
  if (!CheckBitMask(key, mask, 16))
    break;
}

We understand now why the executable never really manages to finish unpacking itself: this loop is extremely unefficient. It comes down to brute-forcing up to $2^{128}$ keys worst-case scenario until finding one that satisifes the mask.

It is much easier to brute-force all the possible subsets of bit indexes given by the mask, since the mask only contains a few one bits.

Finally, let's reverse the DecryptData function.

__int64 __fastcall DecryptData(__int64 data, _BYTE *outbuf, __int64 key, __int64 n_blocks) {
  __int64 result; // rax
  __int64 v5; // rcx
  _BYTE *v6; // r9
  _BYTE *v7; // r8
  __int64 v9; // [rsp+20h] [rbp-30h]
  _BYTE *v11; // [rsp+28h] [rbp-28h]
  __int64 v13; // [rsp+38h] [rbp-18h]
  char v14; // [rsp+40h] [rbp-10h]

  v14 = 1;
  result = InitRoundKeys(key);
  while ( n_blocks ) {
    result = DecryptAESBlock(data, outbuf);
    v5 = 0i64;
    v6 = v11;
    if ( v14 == 1 ) {
      v7 = &mask;
      v14 = 0;
    } else {
      v7 = (_BYTE *)(v9 - 16);
    }
    do {
      *v6++ ^= *v7++;
      ++v5;
    }
    while ( v5 != 16 );
    data = v9 + 16;
    outbuf = v11 + 16;
    n_blocks = v13 - 1;
  }
  return result;
}

The InitRoundKeys and DecryptAESBlock functions were easily recognizable because of the heavy use of XMM registers and AES-related instructions specific to Intel. For instance, the following DecryptAESBlock function leverages the aesdec instruction with the XMM registers xmm5 to xmm15 already filled with the round keys.

As we can see, a XOR is performed at the end of each block, hinting at CBC mode decryption. Under this hypothesis, we notice the mask is also used as the initialization vector for the decryption. One can verify the hypothesis of AES-128-CBC by debugging the executable.

All there is left for us to do is to implement the unpacking algorithm for generalized executables.

Solution implementation

If we ask for a few more executables, we notice the only parts that change through these are the encrypted data and the mask value.

We can then easily fetch an executable, extract the relevant parts (the packed data and the mask) and run our "optimized brute-force" to find the key. Once the binary is unpacked, we get a new executable that runs:

Fortunately for us, there is no obfuscation in this unpacked binary: the secret string can be found inside it in cleartext.

╭─face@0xff ~/ctf/fcsc/reverse/hyperpacker                                                        
╰─$ python3.9 solve_pow.py                                                                        
[+] Opening connection to challenges.france-cybersecurity-challenge.fr on port 2202: Done         
[*] Solving PoW                                                                                   
[+] Solved PoW: 1:26:220507:603e6556f2108faf::kDm1OewN30GkkNXF:6wHth                              
b'Here is your binary:\n'                                                                         
[+] Binary written at /tmp/hyperpacker.bin                                                        
b'Please input the secret within 60 seconds:\n'                                                   
00000000000000190010000000000000                                                                  
b'MZ\x90\x00\x03\x00\x00\x00\x04\x00\x00\x00\xff\xff\x00\x00'                                     
00000000000000190000000000000000                                                                  
b'BLPEN3FICK5OJEXG80YQS83HRIDGAS1M'                                                               
b'Well done!\n'                                                                                

[...]

b'Here is your binary:\n'                                                                         
[+] Binary written at /tmp/hyperpacker.bin                                                        
b'Please input the secret within 60 seconds:\n'                                                   
000000ac7488349c0000000000000000                                                                  
b'MZ\x90\x00\x03\x00\x00\x00\x04\x00\x00\x00\xff\xff\x00\x00'                                     
0000002c300034840000000000000000                                                                  
b'G46PZ0L9PIMUZAUA6J7SK4USE451ZIQK'                                                               
b'Well done!\n'                                                                                   
b'Congratulations! Here is your flag: FCSC{2b60aef3d241d3c37c30373a9a4446017a6d5b6761ae5492e3f824e
280cb8ceb}\n'

Solve script

import subprocess
from pwn import *
from Crypto.Cipher import AES
from itertools import chain, combinations


def powerset(iterable):
    xs = list(iterable)
    return chain.from_iterable(combinations(xs,n) for n in range(len(xs)+1))


def decrypt(data, key, IV):
    cipher = AES.new(key, AES.MODE_CBC, IV=IV)
    return cipher.decrypt(data)


HOST = args.HOST or "challenges.france-cybersecurity-challenge.fr"
PORT = args.PORT or 2202


def main():
    r = remote(HOST, PORT)
    _ = r.recvlines(5)

    cmdline = r.recvline().strip().decode("utf-8").split(" ")
    assert cmdline[0] == "hashcash"
    assert cmdline[1] == "-mb26"
    assert cmdline[2].isalnum()

    log.info(f"Solving PoW")
    solution = subprocess.check_output([cmdline[0], cmdline[1], cmdline[2]])
    log.success(f"Solved PoW: {solution.decode()}")

    r.send(solution)

    while True:
        _ = r.recvline()
        print(_)

        encoded = r.recvline().strip()
        binary = b64d(encoded)

        with open("/tmp/hyperpacker.bin", "wb") as fp:
            fp.write(binary)

        log.success("Binary written at /tmp/hyperpacker.bin")
        print(r.recvline())

        magic_data = binary[0x400:0x400 + 0x48a80]
        IV = binary[0x48e80:0x48e80 + 0x10]
        print(IV.hex())
        
        bin_iv = f"{int.from_bytes(IV, byteorder='big'):0128b}"
        ones = [i for i in range(128) if bin_iv[i] == "1"]

        found = False
        for subset in powerset(ones):
            int_key = 0
            for bit_index in subset:
                int_key ^= 1 << (127 - bit_index)

            key = int_key.to_bytes(length=16, byteorder="big")

            first_block = decrypt(magic_data[:16], key, IV)
            if first_block[:2] == b"MZ":
                if b"PE" in decrypt(magic_data[:256], key, IV):
                    found = True
                    break

        if not found:
            print("not found :(")
            exit(1)

        print(first_block)
        print(key.hex())

        decrypted = decrypt(magic_data, key, IV)
        secret = decrypted.split(b"SECRET is : ")[1].split(b"\x00")[0]
        print(secret)

        r.sendline(secret)
        print(r.recvline())


if __name__ == "__main__":
    main()

"""
FCSC{2b60aef3d241d3c37c30373a9a4446017a6d5b6761ae5492e3f824e280cb8ceb}
"""

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