user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1c0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1c0(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x1b4(%rbp)
  jmp    L30
L97:
  lea    -0x1a0(%rbp),%rdx
  mov    -0x1b4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x1b4(%rbp)
L30:
  mov    -0x1c0(%rbp),%eax
  cmp    %eax,-0x1b4(%rbp)
  jl     L97
  lea    -0x1bc(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x1b0(%rbp)
  jmp    L34
L31:
  lea    -0xd0(%rbp),%rdx
  mov    -0x1b0(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x1b0(%rbp)
L34:
  mov    -0x1bc(%rbp),%eax
  cmp    %eax,-0x1b0(%rbp)
  jl     L31
  movl   $0x0,-0x1b8(%rbp)
  jmp    L135
L12:
  movl   $0x0,-0x1ac(%rbp)
  mov    -0x1b8(%rbp),%eax
  mov    %eax,-0x1a8(%rbp)
  movl   $0x0,-0x1a4(%rbp)
  jmp    L61
L69:
  mov    -0x1a8(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jne    L123
  addl   $0x1,-0x1ac(%rbp)
L123:
  addl   $0x1,-0x1a8(%rbp)
  addl   $0x1,-0x1a4(%rbp)
L61:
  mov    -0x1bc(%rbp),%edx
  mov    -0x1b8(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x1a8(%rbp)
  jge    L13
  mov    -0x1bc(%rbp),%eax
  cmp    %eax,-0x1a4(%rbp)
  jl     L69
L13:
  mov    -0x1bc(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jne    L144
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L52
L144:
  mov    -0x1bc(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  je     L121
  mov    -0x1c0(%rbp),%eax
  mov    -0x1bc(%rbp),%edx
  sub    %edx,%eax
  add    $0x1,%eax
  cltq
  mov    %rax,%rdx
  lea    -0x1a0(%rbp),%rax
  cmp    %rax,%rdx
  jne    L121
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L121:
  addl   $0x1,-0x1b8(%rbp)
L135:
  mov    -0x1c0(%rbp),%eax
  mov    -0x1bc(%rbp),%edx
  sub    %edx,%eax
  add    $0x1,%eax
  cmp    %eax,-0x1b8(%rbp)
  jle    L12
L52:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D1
type: string
name: retrieve_data
arguments: tag: D23
type: string
name: retrieve_data
arguments: tag: D12
type: string

tool: D1: %d

tool: D23: YES

tool: D12: NO

assistant: ```c
int main() {
  int N1, N2, a[50], b[50], k, x, y;
  scanf("%d", &N1);
  for (int i = 0; i < N1; i++) {
    scanf("%d", &a[i]);
  }
  scanf("%d", &N2);
  for (int i = 0; i < N2; i++) {
    scanf("%d", &b[i]);
  }
  for (k = 0; k <= N1 - N2 + 1; k++) {
    int count = 0, x = k, y = 0;
    while ((x < k + N2) && (y <= N2 - 1)) {
      if (a[x] == b[y])
        count++;
      x++;
      y++;
    }
    if (count == N2) {
      printf("YES");
      break;
    } else if ((count != N2) && (a == N1 - N2 + 1))
      printf("NO");
  }
  return 0;
}
```