user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x38(%rbp),%rdx
  lea    -0x3c(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  mov    $0x8,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  mov    $0x8,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    $0x7f,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L47
L71:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L42
L80:
  mov    -0x30(%rbp),%rax
  mov    0x8(%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,(%rdx)
  mov    -0x30(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x30(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
L42:
  mov    -0x38(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x30(%rbp)
  jae    L0
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jl     L80
  jmp    L0
L84:
  mov    -0x28(%rbp),%rax
  mov    0x8(%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,(%rdx)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
L0:
  mov    -0x38(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x28(%rbp)
  jae    L29
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jg     L84
L29:
  addl   $0x1,-0x34(%rbp)
L47:
  mov    -0x3c(%rbp),%eax
  cmp    %eax,-0x34(%rbp)
  jl     L71
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    -0x20(%rbp),%rdx
  add    $0x8,%rdx
  mov    (%rdx),%rdx
  sub    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <isprime@plt>
  test   %rax,%rax
  je     L97
  lea    D44(%rip),%rax
  jmp    L34
L97:
  lea    D45(%rip),%rax
L34:
  mov    -0x10(%rbp),%rdx
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L75
  call   <__stack_chk_fail@plt>
L75:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
type_guessed: string
name: extract_symbolic_data
arguments: label: D14
type_guessed: string
name: extract_symbolic_data
arguments: label: D44
type_guessed: string
name: extract_symbolic_data
arguments: label: D45
type_guessed: string
name: extract_symbolic_data
arguments: label: D20
type_guessed: string

tool: D12: %d %d

tool: D14: %ld

tool: D44: YES

tool: D45: NO

tool: D20: %s
%d


assistant: ```c
int main(void) {
  int n, k;
  scanf("%d %d", &n, &k);
  long *p1, *p2, temp;
  long *min = (long *)calloc(k + 1, sizeof(long));
  long *max = (long *)calloc(k + 1, sizeof(long));
  memset(min, 0x7F, (k + 1) * sizeof(long));
  for (int i = 0; i < n; i++) {
    scanf("%ld", max);
    *(p1 = min) = *(p2 = max);
    while (p1 < min + k && p1[0] < p1[1]) {
      {
        temp = (p1)[1];
        (p1)[1] = *(p1);
        *(p1)++ = temp;
      };
    }
    while (p2 < max + k && p2[0] > p2[1]) {
      {
        temp = (p2)[1];
        (p2)[1] = *(p2);
        *(p2)++ = temp;
      };
    }
  }
  temp = max[1] - min[1];
  printf("%s\n%d\n", isprime(temp) ? "YES" : "NO", temp);
  free(min);
  free(max);
  return 0;
}
```