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

```asm
<main>:
  endbr64
  push   %r13
  lea    D4(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L79
  xor    %ebp,%ebp
  nopl   (%rax)
L49:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L87:
  mov    $0x2a,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebx,%ebp
  jge    L87
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    %ebp,0x4(%rsp)
  jg     L49
L79:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%ecx
  test   %ecx,%ecx
  jle    L89
  xor    %ebp,%ebp
  lea    D46(%rip),%r12
  nop
L75:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L111:
  add    $0x1,%ebx
  xor    %eax,%eax
  mov    %r12,%rdi
  mov    %ebx,%esi
  call   <printf@plt>
  cmp    %ebp,%ebx
  jle    L111
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    %ebp,0x4(%rsp)
  jg     L75
L89:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L50
  xor    %ebp,%ebp
  lea    D7(%rip),%r12
  nopl   0x0(%rax)
L51:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L115:
  lea    0x41(%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jge    L115
  mov    $0xa,%edi
  add    $0x1,%ebp
  call   <putchar@plt>
  cmp    %ebp,0x4(%rsp)
  jg     L51
L50:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  js     L44
  mov    $0x1,%ebp
  xor    %r12d,%r12d
  lea    D5(%rip),%r13
L81:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L56:
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  add    $0x1,%ebp
  cmp    %ebx,%r12d
  jge    L56
  mov    $0xa,%edi
  add    $0x1,%r12d
  call   <putchar@plt>
  cmp    %r12d,0x4(%rsp)
  jge    L81
L44:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type: string
name: retrieve_data
arguments: data_tag: D11
type: string
name: retrieve_data
arguments: data_tag: D46
type: string
name: retrieve_data
arguments: data_tag: D7
type: string
name: retrieve_data
arguments: data_tag: D5
type: string

tool: D4: Visina?

tool: D11: %d

tool: D46: %d 

tool: D7: %c 

tool: D5: %2d 

assistant: ```c
int main() {
  int v;
  printf("Visina?");
  scanf("%d", &v);
  int niz[v], i;
  for (i = 0; i < v; i++) {
    int j;
    for (j = 0; j <= i; j++)
      printf("*");
    printf("\n");
  }
  printf("\n");
  for (i = 0; i < v; i++) {
    int j;
    for (j = 0; j <= i; j++)
      printf("%d ", j + 1);
    printf("\n");
  }
  printf("\n");
  for (i = 0; i < v; i++) {
    int j;
    for (j = 0; j <= i; j++)
      printf("%c ", 'A' + j);
    printf("\n");
  }
  printf("\n");
  int k = 1;
  for (i = 0; i <= v; i++) {
    int j;
    for (j = 0; j <= i; ++j) {
      printf("%2d ", k++);
    }
    printf("\n");
  }
  printf("\n");
  return 0;
}
```