user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebp
  cmp    $0x2,%ebp
  jle    L181
  lea    -0x3(%rbp,%rbp,1),%edi
  mov    $0x1,%ecx
  mov    $0x0,%esi
  jmp    L125
L146:
  add    $0x1,%eax
  cmp    %eax,%edx
  jne    L146
  lea    (%rcx,%rsi,1),%esi
L199:
  add    $0x2,%ecx
  cmp    %ecx,%edi
  je     L5
L125:
  mov    %ecx,%edx
  mov    $0x0,%eax
  test   %ecx,%ecx
  jg     L146
  jmp    L199
L5:
  sub    $0x1,%ebp
L106:
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L45
  mov    $0x0,%ebx
L219:
  mov    $0x20,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jg     L219
L45:
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebp,%eax
  jle    L216
L130:
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebp,%eax
  jg     L130
L216:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%ebx
  cmp    $0x2,%ebx
  jle    L64
  mov    $0x1,%r12d
  mov    $0x1,%ebp
L173:
  sub    %ebp,%ebx
  sub    $0x1,%ebx
  je     L171
L213:
  mov    $0x20,%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  jne    L213
L171:
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    %r12d,%r13d
  test   %r12d,%r12d
  jle    L158
  mov    $0x0,%ebx
L86:
  mov    $0x23,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %ebx,%r13d
  jne    L86
L158:
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    %ebp,%ebx
  not    %ebx
  add    0x4(%rsp),%ebx
  je     L111
L52:
  mov    $0x20,%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  jne    L52
L111:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%ebp
  mov    0x4(%rsp),%ebx
  add    $0x2,%r12d
  lea    -0x1(%rbx),%eax
  cmp    %ebp,%eax
  jg     L173
L64:
  mov    0x4(%rsp),%eax
  add    %eax,%eax
  cmp    $0x1,%eax
  jle    L119
  mov    $0x0,%ebx
L210:
  mov    $0x2a,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax,%rax,1),%eax
  cmp    %ebx,%eax
  jg     L210
L119:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L57
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L181:
  mov    $0x0,%esi
  mov    $0x1,%ebp
  jmp    L106
L57:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D9
datatype: string
name: unpack_data
arguments: data_tag: D18
datatype: string

tool: D9: %d

tool: D18: %d


assistant: ```c
int main() {
  int i, n, k, l;
  l = 0;
  scanf("%d", &n);
  for (k = 1; k < n - 1; k++)
    for (i = 0; i < k * 2 - 1; i++) {
      l++;
    }
  printf("%d\n", l);
  for (i = 0; i < n - 1; i++)
    printf(" ");
  printf("*");
  for (i = 0; k < n - 1; i++)
    printf(" ");
  printf("\n");
  for (k = 1; k < n - 1; k++) {
    for (i = n - k - 1; i != 0; i--)
      printf(" ");
    { printf("*"); }
    for (i = 0; i < k * 2 - 1; i++)
      printf("#");
    { printf("*"); }
    for (i = n - k - 1; i != 0; i--)
      printf(" ");
    { printf("\n"); }
  }
  for (i = 0; i < n * 2 - 1; i++) {
    printf("*");
  }
  return 0;
}
```