user: What is the c source code of the assembly code below?

```asm
<Problem7>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xf68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1f58(%rsp)
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <puts@plt>
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D48(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  mov    $0x0,%ebx
  mov    $0x2,%esi
  jmp    L167
L181:
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L40
  movslq %ebx,%rax
  mov    %esi,0x10(%rsp,%rax,4)
  add    $0x1,%ebx
L40:
  add    $0x1,%esi
  cmp    $0x186a0,%esi
  je     L11
L167:
  cmp    $0x3,%esi
  jle    L181
  test   $0x1,%sil
  je     L40
  mov    $0x2,%ecx
L164:
  add    $0x1,%ecx
  mov    %esi,%eax
  cltd
  idiv   %ecx
  cmp    %ecx,%eax
  jl     L181
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L164
  jmp    L40
L11:
  cmp    $0x2,%edi
  jle    L112
  lea    0xfb0(%rsp),%rdx
  mov    $0x2,%eax
L10:
  mov    %eax,(%rdx)
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %eax,%edi
  jne    L10
L112:
  mov    %edi,%esi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x2,0xc(%rsp)
  jle    L13
  lea    0x10(%rsp),%r15
  lea    -0x1(%rbx),%ebp
  shl    $0x2,%rbp
  lea    0x14(%rsp,%rbp,1),%r12
  add    %r15,%rbp
  mov    $0x0,%r13d
  lea    D6(%rip),%r14
  jmp    L130
L34:
  add    $0x4,%rcx
  cmp    %rcx,%r12
  je     L141
L44:
  mov    %esi,%eax
  cltd
  idivl  (%rcx)
  test   %edx,%edx
  je     L141
  cmp    %rbp,%rcx
  jne    L34
  mov    0xc(%rsp),%edx
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L141:
  add    $0x1,%r13
  mov    0xc(%rsp),%eax
  sub    $0x2,%eax
  cmp    %r13d,%eax
  jle    L13
L130:
  test   %ebx,%ebx
  jle    L141
  mov    0xfb0(%rsp,%r13,4),%esi
  mov    %r15,%rcx
  jmp    L44
L13:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x1f58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L128
  add    $0x1f68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L128:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D4
datatype: string
name: disassemble_data
arguments: tag: D28
datatype: string
name: disassemble_data
arguments: tag: D48
datatype: string
name: disassemble_data
arguments: tag: D23
datatype: string
name: disassemble_data
arguments: tag: D6
datatype: string

tool: D4: Problem 7: Given a number n, all the fractions less than 1, such that n is the denominator, is printed (in lowest terms).

tool: D28: Enter n: 

tool: D48: %d

tool: D23: The output is: 1/%d

tool: D6: , %d/%d

assistant: ```c
void Problem7() {
  int n, temp, i, j, arr[1000], k = 0, ix, jx, tempo[1000], tempor;
  printf("Problem 7: Given a number n, all the fractions less than 1, such that n is the denominator, is printed (in lowest terms).\n");
  printf("Enter n: ");
  scanf("%d", &n);
  for (i = 2; i < 100000; i++) {
    for (j = 2; j <= (i / j); j++) {
      if (!(i % j)) {
        break;
      }
    }
    if (j > (i / j)) {
      if (n % i == 0) {
        arr[k] = i;
        k++;
      }
    }
  }
  temp = k;
  tempor = 2;
  for (i = 0; i < n - 2; i++) {
    tempo[i] = tempor;
    tempor++;
  }
  printf("The output is: 1/%d", n);
  for (i = 0; i < n - 2; i++) {
    for (k = 0; k < temp; k++) {
      if (tempo[i] % arr[k] == 0) {
        break;
      }
      if (k + 1 == temp) {
        printf(", %d/%d", tempo[i], n);
      }
    }
  }
  printf("\n");
}
```