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

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x32,%ecx
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  lea    D49(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x4f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4e8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  lea    0x1b0(%rsp),%rdx
  mov    %rbp,%rdi
  lea    0x350(%rsp),%rbx
  lea    0xc(%rsp),%rsi
  lea    0x1b4(%rsp),%r14
  rep stos %rax,%es:(%rdi)
  mov    $0x32,%ecx
  movl   $0x0,(%rdi)
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    $0x32,%ecx
  movl   $0x0,(%rdi)
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  lea    D32(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L206
L84:
  mov    %r12,%rdi
  xor    %eax,%eax
  lea    (%rbx,%r13,4),%rdx
  mov    %r14,%rsi
  call   <scanf@plt>
  mov    (%r14),%ecx
  mov    (%rbx,%r13,4),%edi
  movdqa D21(%rip),%xmm1
  cmp    %edi,%ecx
  jg     L209
  mov    %edi,%eax
  sub    %ecx,%eax
  lea    0x1(%rax),%r8d
  cmp    $0x2,%eax
  jbe    L103
  movslq %ecx,%rax
  lea    0x0(%rbp,%rax,4),%rdx
  mov    %r8d,%eax
  shr    $0x2,%eax
  lea    -0x1(%rax),%esi
  lea    0x10(%rdx),%rax
  shl    $0x4,%rsi
  add    %rax,%rsi
  jmp    L91
  nop
L98:
  add    $0x10,%rax
L91:
  movdqu (%rdx),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rdx)
  mov    %rax,%rdx
  cmp    %rsi,%rax
  jne    L98
  mov    %r8d,%eax
  and    $0xfffffffc,%eax
  add    %eax,%ecx
  cmp    %r8d,%eax
  je     L209
L103:
  movslq %ecx,%rax
  addl   $0x1,0x10(%rsp,%rax,4)
  lea    0x1(%rcx),%eax
  cmp    %edi,%eax
  jg     L209
  cltq
  add    $0x2,%ecx
  addl   $0x1,0x10(%rsp,%rax,4)
  cmp    %edi,%ecx
  jg     L209
  movslq %ecx,%rcx
  addl   $0x1,0x10(%rsp,%rcx,4)
L209:
  mov    0xc(%rsp),%eax
  add    $0x1,%r13
  add    $0x4,%r14
  cmp    %r13d,%eax
  jge    L84
  test   %eax,%eax
  jle    L206
  mov    0x10(%rsp),%ecx
  lea    0x1(%rax),%edi
  xor    %r12d,%r12d
  mov    $0x1,%eax
  nopl   (%rax)
L123:
  mov    0x0(%rbp,%rax,4),%edx
  cmp    %eax,%edx
  jne    L141
  cmp    %edx,%ecx
  setle  %sil
  test   %sil,%sil
  cmovne %edx,%ecx
  cmovne %eax,%r12d
L141:
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L123
L160:
  mov    %r12d,%esi
  lea    D17(%rip),%rdi
  lea    D13(%rip),%rbp
  call   <printf@plt>
  mov    0xc(%rsp),%edx
  mov    $0x1,%esi
  test   %edx,%edx
  jle    L113
  nopl   0x0(%rax)
L0:
  movslq %esi,%rax
  mov    0x1b0(%rsp,%rax,4),%ebx
  mov    0x350(%rsp,%rax,4),%eax
  cmp    %eax,%ebx
  jg     L216
  nopl   (%rax)
L60:
  cmp    %r12d,%ebx
  je     L41
  add    $0x1,%ebx
  cmp    %eax,%ebx
  jle    L60
L216:
  add    $0x1,%esi
  cmp    %edx,%esi
  jle    L0
L113:
  mov    0x4e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L73
  add    $0x4f0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L41:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  movslq 0xc(%rsp),%rax
  mov    %rax,%rsi
  mov    %eax,%edx
  mov    0x350(%rsp,%rax,4),%eax
  cmp    %eax,%ebx
  jle    L60
  add    $0x1,%esi
  cmp    %edx,%esi
  jle    L0
  jmp    L113
L206:
  xor    %r12d,%r12d
  jmp    L160
L73:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D49
datatype: string
name: unpack_data
arguments: data_label: D32
datatype: string
name: unpack_data
arguments: data_label: D21
datatype: byte[16]
name: unpack_data
arguments: data_label: D17
datatype: string
name: unpack_data
arguments: data_label: D13
datatype: string

tool: D49: %d %d

tool: D32: %d

tool: {"D21": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D17: Numero de guardioes sinceros = %d


tool: D13: Guardiao %d


assistant: ```c
int main() {
  int n, g[101], i, j, min[101], max[101], nSinc = 0;
  for (i = 0; i < 101; i++) {
    g[i] = 0;
    min[i] = 0;
    max[i] = 0;
  }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &min[i], &max[i]);
    for (j = min[i]; j <= max[i]; j++)
      g[j]++;
  }
  for (i = 1; i <= n; i++) {
    if (g[i] >= g[nSinc] && i == g[i])
      nSinc = i;
  }
  printf("Numero de guardioes sinceros = %d\n", nSinc);
  for (i = 1; i <= n; i++) {
    for (j = min[i]; j <= max[i]; j++)
      if (j == nSinc) {
        printf("Guardiao %d\n", i);
        i = n;
      }
  }
  return 0;
}
```