user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  lea    D24(%rip),%rsi
  lea    D33(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D18(%rip),%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  mov    %rax,0x8(%rsp)
  lea    0x20(%rsp),%rax
  mov    %rax,0x18(%rsp)
  nopl   0x0(%rax,%rax,1)
L46:
  mov    0x18(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    0x20(%rsp),%edx
  test   %edx,%edx
  je     L41
  mov    0x8(%rsp),%rdi
  lea    D54(%rip),%rsi
  xor    %eax,%eax
  call   <fprintf@plt>
  mov    0x20(%rsp),%eax
  test   %eax,%eax
  jle    L46
  movl   $0x0,0x14(%rsp)
  lea    0x24(%rsp),%r15
  lea    0x2e(%rsp),%r14
  nopw   0x0(%rax,%rax,1)
L40:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rbx
  cltd
  imul   $0x4ec4ec4f,%rbx,%rbx
  sar    $0x23,%rbx
  sub    %edx,%ebx
  imul   $0x1a,%ebx,%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  nop
L25:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%r12
  cltd
  imul   $0x4ec4ec4f,%r12,%r12
  sar    $0x23,%r12
  sub    %edx,%r12d
  imul   $0x1a,%r12d,%edx
  sub    %edx,%eax
  mov    %eax,%r12d
  cmp    %eax,%ebx
  jl     L25
  mov    0x8(%rsp),%rdi
  mov    %eax,%ecx
  mov    %ebx,%edx
  xor    %eax,%eax
  lea    D66(%rip),%rsi
  call   <fprintf@plt>
  test   %r12d,%r12d
  jle    L13
  xor    %r13d,%r13d
  mov    %r12d,0x10(%rsp)
  mov    %r13d,%r12d
  nopl   (%rax)
L48:
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <rand@plt>
  cltd
  idiv   %ebx
  xor    %eax,%eax
  mov    %edx,%r13d
  call   <rand@plt>
  lea    0x61(%r13),%ecx
  mov    $0xa,%esi
  mov    %r15,%rdi
  cltd
  idiv   %ebx
  mov    %edx,0x4(%rsp)
  mov    %rbp,%rdx
  call   <snprintf@plt>
  mov    0x4(%rsp),%edx
  mov    $0xa,%esi
  mov    %r14,%rdi
  lea    0x61(%rdx),%ecx
  mov    %rbp,%rdx
  call   <snprintf@plt>
  mov    0x8(%rsp),%rdi
  mov    %r14,%rcx
  mov    %r15,%rdx
  lea    D7(%rip),%rsi
  xor    %eax,%eax
  call   <fprintf@plt>
  cmp    0x10(%rsp),%r12d
  jne    L48
L13:
  addl   $0x1,0x14(%rsp)
  mov    0x14(%rsp),%eax
  cmp    %eax,0x20(%rsp)
  jg     L40
  jmp    L46
L41:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L67
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L67:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D24
type_infered: string
name: unpack_data
arguments: data_label: D33
type_infered: string
name: unpack_data
arguments: data_label: D18
type_infered: string
name: unpack_data
arguments: data_label: D3
type_infered: string
name: unpack_data
arguments: data_label: D54
type_infered: string
name: unpack_data
arguments: data_label: D66
type_infered: string
name: unpack_data
arguments: data_label: D7
type_infered: string

tool: D24: w

tool: D33: componentes.txt

tool: D18: %c

tool: D3: %d

tool: D54: %d


tool: D66: %d %d


tool: D7: %s %s


assistant: ```c
int main() {
  FILE *out = fopen("componentes.txt", "w");
  char b1[10], b2[10];
  int n, v, e, vv, ee;
  while (1) {
    scanf("%d", &n);
    if (n == 0)
      break;
    fprintf(out, "%d\n", n);
    for (int i = 0; i < n; i++) {
      v = (rand() % 26);
      e = v + 1;
      while (e > v)
        e = (rand() % 26);
      fprintf(out, "%d %d\n", v, e);
      for (int j = 0; j < e; j++) {
        vv = (rand() % v);
        ee = (rand() % v);
        snprintf(b1, 10, "%c", vv + 97);
        snprintf(b2, 10, "%c", ee + 97);
        fprintf(out, "%s %s\n", b1, b2);
      }
    }
  }
}
```