user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  lea    D7(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1e8(%rsp)
  xor    %eax,%eax
  lea    0x3c(%rsp),%rax
  lea    0x40(%rsp),%rdx
  mov    %rax,0x20(%rsp)
  mov    %rax,%rsi
  xor    %eax,%eax
  lea    0x50(%rsp),%r15
  mov    %rdx,0x28(%rsp)
  call   <scanf@plt>
  mov    0x3c(%rsp),%eax
  or     0x40(%rsp),%eax
  movl   $0x0,0x1c(%rsp)
  je     L19
L76:
  addl   $0x1,0x1c(%rsp)
  mov    0x1c(%rsp),%esi
  lea    D55(%rip),%rdi
  xor    %ebx,%ebx
  lea    0x50(%rsp),%r12
  lea    D53(%rip),%rbp
  call   <printf@plt>
  mov    0x3c(%rsp),%ecx
  test   %ecx,%ecx
  jle    L16
  nopl   0x0(%rax,%rax,1)
L128:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <scanf@plt>
  add    $0x4,%r12
  cmp    %ebx,0x3c(%rsp)
  jg     L128
L16:
  lea    0x48(%rsp),%rax
  mov    0x40(%rsp),%edx
  movl   $0x0,0x18(%rsp)
  lea    D53(%rip),%r13
  mov    %rax,0x10(%rsp)
  lea    0x44(%rsp),%rax
  mov    %rax,0x8(%rsp)
  test   %edx,%edx
  jle    L100
  nopl   0x0(%rax)
L30:
  mov    0x10(%rsp),%rdx
  mov    0x8(%rsp),%rsi
  xor    %eax,%eax
  xor    %ebx,%ebx
  lea    D7(%rip),%rdi
  xor    %ebp,%ebp
  lea    0x4c(%rsp),%r12
  call   <scanf@plt>
  mov    0x44(%rsp),%eax
  test   %eax,%eax
  jg     L34
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %eax,%ebp
L84:
  add    $0x1,%ebx
  cmp    %r14d,%ebx
  jge    L33
L34:
  mov    %r12,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x48(%rsp),%ecx
  mov    0x44(%rsp),%r14d
  lea    0x1(%rbp),%eax
  cmp    %ecx,0x4c(%rsp)
  je     L6
  cmp    %r14d,%eax
  jge    L84
  lea    -0x2(%r14),%edx
  cltq
  add    $0x1,%ebx
  sub    %ebp,%edx
  lea    (%r15,%rax,4),%rsi
  lea    -0x4(%r15,%rax,4),%rdi
  lea    0x4(,%rdx,4),%rdx
  call   <memmove@plt>
  cmp    %r14d,%ebx
  jl     L34
L33:
  addl   $0x1,0x18(%rsp)
  mov    0x18(%rsp),%eax
  cmp    %eax,0x40(%rsp)
  jg     L30
L100:
  mov    0x50(%rsp),%esi
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rdx
  xor    %eax,%eax
  mov    0x20(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  mov    0x3c(%rsp),%eax
  or     0x40(%rsp),%eax
  jne    L76
L19:
  mov    0x1e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x1f8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D7
type_infered: string
name: parse_data
arguments: data_label: D55
type_infered: string
name: parse_data
arguments: data_label: D53
type_infered: string
name: parse_data
arguments: data_label: D38
type_infered: string

tool: D7: %d %d

tool: D55: Teste %d


tool: D53: %d

tool: D38: %d



assistant: ```c
int main() {
  int teste = 0;
  int participantes, rodadas;
  int part[100];
  int npart;
  int ordem;
  int acao;
  int posacao;
  int i, j, k;
  scanf("%d %d", &participantes, &rodadas);
  while (participantes || rodadas) {
    printf("Teste %d\n", ++teste);
    for (i = 0; i < participantes; i++)
      scanf("%d", &(part[i]));
    for (i = 0; i < rodadas; i++) {
      scanf("%d %d", &npart, &ordem);
      posacao = 0;
      for (j = 0; j < npart; j++) {
        scanf("%d", &acao);
        if (acao == ordem)
          posacao++;
        else
          for (k = posacao + 1; k < npart; k++)
            part[k - 1] = part[k];
      }
    }
    printf("%d\n\n", part[0]);
    scanf("%d %d", &participantes, &rodadas);
  }
  return 0;
}
```