(10 Points) Problem 5- Reverse Engineering Reverse engineer (disassemble) the following code snippet into an equivalent C code snippet. mov ebx, 0ah xor ecx, ecx someLoop: xor eax, eax push eax push ecx push 3 push 0ah push eax call cybr mov value, eax add esp, 14h p eax, ebx jle someLoop mul ecx, eax