Problem 1: Consider the following assembly code:

.LC0 .string "ans %d " main: .LFB0: pushq %rbp movq %rsp, %rbp subq $16, %rsp movabsq $83162457113523645, %rax movq %rax, -8(%rbp) movl $0, -12(%rbp) jmp .L2 .L3: movq -8(%rbp), %rbx movl %ebx, %ecx andl $1, %ecx movl -12(%rbp), %ebx addl %ecx, %ebx movl %ebx, -12(%rbp) sarq -8(%rbp) .L2: cmpq $0, -8(%rbp) jg .L3 movl -12(%rbp), %ebx movl %ebx, %esi movl $.LC0, %edi movl $0, %ebx call printf leave ret 

This code came from the following skeleton C file, and optimized with O0, so gcc -O0 -S was the exact command to compile the file. Complete the below given C code using the above information . Step 1 might be to ignore the skeleton file and create a C file from the assembly code, and then rewrite the code to fit the skeleton file.

#include  int main() { long int val=_________; int result=_________; for(;val>_________;val=_________) result += ___________; printf("ans %d ",result); }