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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x4000(%rsp),%r11
L75:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L75
  sub    $0xe50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L76
L9:
  movl   $0x0,-0x4e40(%rbp)
  lea    -0x4e30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4e3c(%rbp)
  lea    -0x4e30(%rbp),%rax
  mov    %rax,%rdi
  call   <ispalindrome@plt>
  test   %rax,%rax
  je     L128
  lea    -0x4e30(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L76
L128:
  mov    -0x4e3c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4e48(%rbp)
  jmp    L79
L68:
  mov    -0x4e48(%rbp),%eax
  cltq
  movzbl -0x4e30(%rbp,%rax,1),%edx
  mov    -0x4e48(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movzbl -0x4e30(%rbp,%rax,1),%eax
  cmp    %al,%dl
  jne    L113
  addl   $0x1,-0x4e40(%rbp)
  subl   $0x1,-0x4e48(%rbp)
L79:
  cmpl   $0x0,-0x4e48(%rbp)
  jns    L68
  jmp    L17
L113:
  nop
L17:
  mov    -0x4e40(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x4e3c(%rbp),%eax
  sub    -0x4e40(%rbp),%eax
  mov    %eax,-0x4e38(%rbp)
  mov    -0x4e38(%rbp),%edx
  lea    -0x4e30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <lengthtobeadd@plt>
  mov    %eax,-0x4e34(%rbp)
  mov    -0x4e34(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4e44(%rbp)
  mov    -0x4e34(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L31
  mov    -0x4e34(%rbp),%eax
  add    %eax,%eax
  sub    %eax,-0x4e3c(%rbp)
  jmp    L21
L31:
  mov    -0x4e3c(%rbp),%eax
  sub    -0x4e34(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4e34(%rbp),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  sar    %eax
  neg    %eax
  add    %edx,%eax
  sub    $0x1,%eax
  mov    %eax,-0x4e3c(%rbp)
L21:
  mov    -0x4e3c(%rbp),%eax
  mov    %eax,-0x4e48(%rbp)
  jmp    L47
L108:
  mov    -0x4e48(%rbp),%eax
  cltq
  movzbl -0x4e30(%rbp,%rax,1),%edx
  mov    -0x4e44(%rbp),%eax
  cltq
  mov    %dl,-0x2720(%rbp,%rax,1)
  addl   $0x1,-0x4e44(%rbp)
  subl   $0x1,-0x4e48(%rbp)
L47:
  cmpl   $0x0,-0x4e48(%rbp)
  jns    L108
  mov    -0x4e44(%rbp),%eax
  cltq
  movb   $0x0,-0x2720(%rbp,%rax,1)
  mov    -0x4e48(%rbp),%eax
  cltq
  movb   $0x0,-0x4e30(%rbp,%rax,1)
  lea    -0x2720(%rbp),%rdx
  lea    -0x4e30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  lea    -0x4e30(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L76:
  lea    -0x4e30(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L9
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L101
  call   <__stack_chk_fail@plt>
L101:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D19
type_guessed: string
name: read_data
arguments: tag: D24
type_guessed: string

tool: D19: %d


tool: D24: %s

assistant: ```c
int main() {
  int i, j, k, l, m, len;
  char a[10000], b[10000];
  while (scanf("%s", a) == 1) {
    l = 0;
    len = strlen(a);
    if (ispalindrome(a))
      printf("%s\n", a);
    else {
      for (i = len - 1; i >= 0; i--) {
        if (a[i] == a[i - 1]) {
          l++;
        } else
          break;
      }
      printf("%d\n", l);
      m = len - l;
      j = lengthtobeadd(a, m);
      printf("%d\n", j);
      k = 0;
      if (j % 2 == 0) {
        len = len - 2 * j;
      } else {
        len = len - j - (j / 2) - 1;
      }
      for (i = len; i >= 0; i--) {
        b[k] = a[i];
        k++;
      }
      b[k] = '\0';
      a[i] = '\0';
      strcat(a, b);
      printf("%s\n", a);
    }
  }
  return 0;
}
```