user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xeb0,%rsp
  lea    D29(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x3ea8(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  cmpq   $0x0,0x8(%rsp)
  jle    L61
  xor    %ebx,%ebx
  lea    0x18(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    D13(%rip),%r12
  lea    D9(%rip),%rbp
  nopl   0x0(%rax)
L221:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %r14,%rdx
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%rsi
  xor    %r9d,%r9d
  xor    %edi,%edi
  mov    $0x2,%ecx
  cmp    $0x1,%rsi
  jle    L84
  nopl   0x0(%rax)
L205:
  mov    %rsi,%rax
  lea    0x20(%rsp),%r8
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L222
  nopw   0x0(%rax,%rax,1)
L214:
  mov    %rsi,%rax
  mov    %rcx,(%r8,%rdi,8)
  add    $0x1,%rdi
  cqto
  idiv   %rcx
  cqto
  mov    %rax,%rsi
  idiv   %rcx
  test   %rdx,%rdx
  je     L214
  mov    $0x1,%r9d
L222:
  cmp    %rcx,%rsi
  jl     L109
  add    $0x1,%rcx
  jmp    L205
  xchg   %ax,%ax
L109:
  cmp    $0x1,%rsi
  jle    L204
  mov    $0x2,%ecx
  jmp    L205
  nopl   (%rax)
L204:
  test   %r9b,%r9b
  je     L186
  mov    %rsi,0x10(%rsp)
L186:
  mov    0x18(%rsp),%rsi
  cmp    $0x1,%rsi
  jle    L141
L116:
  xor    %r10d,%r10d
  xor    %r8d,%r8d
  mov    $0x2,%ecx
L57:
  mov    %rsi,%rax
  lea    0x1f60(%rsp),%r9
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L58
  nopl   (%rax)
L151:
  mov    %rsi,%rax
  mov    %rcx,(%r9,%r8,8)
  add    $0x1,%r8
  cqto
  idiv   %rcx
  cqto
  mov    %rax,%rsi
  idiv   %rcx
  test   %rdx,%rdx
  je     L151
  mov    $0x1,%r10d
L58:
  cmp    %rcx,%rsi
  jl     L234
  add    $0x1,%rcx
  jmp    L57
  cs nopw 0x0(%rax,%rax,1)
L234:
  cmp    $0x1,%rsi
  jle    L239
  mov    $0x2,%ecx
  jmp    L57
  nopl   (%rax)
L239:
  test   %r10b,%r10b
  je     L141
  mov    %rsi,0x18(%rsp)
L141:
  test   %rdi,%rdi
  je     L147
  xor    %eax,%eax
  mov    $0x1,%esi
  lea    0x20(%rsp),%r8
  lea    0x1f60(%rsp),%r9
  nopw   0x0(%rax,%rax,1)
L176:
  mov    (%r8,%rax,8),%rdx
  cmp    (%r9,%rax,8),%rdx
  jne    L2
  imul   %rdx,%rsi
L2:
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L176
L168:
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    %rbx,0x8(%rsp)
  jg     L221
L61:
  mov    0x3ea8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L60
  add    $0x3eb0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L84:
  mov    0x18(%rsp),%rsi
  cmp    $0x1,%rsi
  jg     L116
L147:
  mov    $0x1,%esi
  jmp    L168
L60:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D29
data_type: string
name: parse_data
arguments: tag: D13
data_type: string
name: parse_data
arguments: tag: D9
data_type: string

tool: D29: %lld

tool: D13: %lld%lld

tool: D9: 
%lld

assistant: ```c
int main() {
  long long int n, i, j = 2, k = 2, l = 0, m = 0, a[1000], b[1000], p = 1, c, d, o;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld%lld", &c, &d);
    while (c > 1) {
      while (c % j == 0) {
        if (c % j == 0) {
          a[l] = j;
          c = c / j;
          l++;
        }
      }
      if (c < j)
        j = 1;
      j++;
    }
    while (d > 1) {
      while (d % k == 0) {
        if (d % k == 0) {
          b[m] = k;
          d = d / k;
          m++;
        }
      }
      if (d < k)
        k = 1;
      k++;
    }
    for (o = 0; o < l; o++) {
      if (a[o] == b[o])
        p = p * a[o];
    }
    printf("\n%lld", p);
    printf("\n");
    j = 2;
    k = 2;
    l = 0;
    m = 0;
    p = 1;
  }
  return 0;
}
```