user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  test   %eax,%eax
  jns    L51
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L77
L51:
  mov    -0x1c(%rbp),%eax
  test   %eax,%eax
  jne    L22
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L77
L22:
  movl   $0x3,-0x18(%rbp)
  jmp    L43
L26:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  addl   $0x1,-0x18(%rbp)
L43:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jle    L26
  mov    -0x1c(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  sub    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jge    L66
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L77
L66:
  mov    -0x1c(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  sub    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jle    L23
  mov    -0x18(%rbp),%eax
  lea    -0x2(%rax),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L77
L23:
  mov    -0x1c(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  sub    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jne    L77
  mov    -0x18(%rbp),%eax
  lea    -0x2(%rax),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L77:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L49
  call   <__stack_chk_fail@plt>
L49:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D18
type_infered: string
name: parse_data
arguments: data_tag: D49
type_infered: string
name: parse_data
arguments: data_tag: D26
type_infered: string
name: parse_data
arguments: data_tag: D22
type_infered: string

tool: D18: Please enter a Fibonacci number: 

tool: D49: %d

tool: D26: 1, 1

tool: D22: %d, %d

assistant: ```c
int main() {
  int number = 0, fibonacciInput = 0, fibonacciA = 1, fibonacciB = 1, fibonacciC = 0;
  printf("Please enter a Fibonacci number: ");
  scanf("%d", &fibonacciInput);
  if (fibonacciInput < 0) {
    printf("1, 1\n");
  } else if (fibonacciInput == 0) {
    printf("1, 1\n");
  } else {
    for (number = 3; fibonacciC <= fibonacciInput; number++) {
      fibonacciC = fibonacciA + fibonacciB;
      fibonacciA = fibonacciB;
      fibonacciB = fibonacciC;
    }
    if ((fibonacciC - fibonacciInput) < (fibonacciInput - fibonacciA)) {
      printf("%d, %d", fibonacciC, number - 1);
    } else if ((fibonacciC - fibonacciInput) > (fibonacciInput - fibonacciA)) {
      printf("%d, %d", fibonacciA, number - 2);
    } else if ((fibonacciC - fibonacciInput) == (fibonacciInput - fibonacciA)) {
      printf("%d, %d", fibonacciA, number - 2);
    }
  }
  return 0;
}
```