user: Converte this assembly snippet into its equivalent C code:

```asm
<six>:
  endbr64
  push   %r15
  mov    $0x31,%esi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  sub    $0x38,%rsp
  movdqa D21(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x676e,%eax
  movl   $0x61697872,0x20(%rsp)
  lea    0xd(%rsp),%r14
  movaps %xmm0,(%rsp)
  movdqa D43(%rip),%xmm0
  mov    %r14,%r13
  lea    0x27(%rsp),%r12
  mov    %ax,0x24(%rsp)
  movb   $0x0,0x26(%rsp)
  movaps %xmm0,0x10(%rsp)
L77:
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    -0xc(%r13),%rbx
  call   <printf@plt>
  nopl   0x0(%rax)
L5:
  movsbl (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r13,%rbx
  jne    L5
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    %r12,%rbx
  je     L50
  movsbl 0x0(%r13),%esi
  add    $0xd,%r13
  jmp    L77
L50:
  mov    %rsp,%r13
  lea    0x1(%rsp),%rbx
  mov    $0x1,%r15d
  mov    %r13,%rdi
  sub    %ebx,%r15d
  call   <puts@plt>
  mov    %rbx,%rdi
  call   <puts@plt>
  movsbl 0x2(%rsp),%esi
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    %r13,%rdi
  call   <puts@plt>
  movsbl (%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  cs nopw 0x0(%rax,%rax,1)
L9:
  movsbl (%rbx),%esi
  add    $0x1,%rbx
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    (%r15,%rbx,1),%eax
  imul   $0xc4ec4ec5,%eax,%eax
  cmp    $0x13b13b13,%eax
  ja     L23
  mov    $0xa,%edi
  call   <putchar@plt>
L23:
  cmp    %r12,%rbx
  jne    L9
  movsbl (%rsp),%esi
  test   %sil,%sil
  je     L61
  nopw   0x0(%rax,%rax,1)
L25:
  add    $0x1,%r13
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movsbl 0x0(%r13),%esi
  test   %sil,%sil
  jne    L25
L61:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r14,%rdi
  call   <puts@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D3
type_infered: string
name: read_data
arguments: label: D21
type_infered: byte[16]
name: read_data
arguments: label: D43
type_infered: byte[16]
name: read_data
arguments: label: D18
type_infered: string
name: read_data
arguments: label: D0
type_infered: string

tool: D3:  %c 

tool: {"D21": [0x31,0x7a,0x68,0x61,0x6e,0x67,0x72,0x78,0x69,0x61,0x6e,0x67,0x00,0x32,0x7a,0x68]}

tool: {"D43": [0x61,0x6e,0x67,0x72,0x78,0x69,0x61,0x6e,0x67,0x00,0x33,0x7a,0x68,0x61,0x6e,0x67]}

tool: D18: %c


tool: D0: --------------------------

assistant: ```c
void six() {
  char str[3][13] = {
      {'1', 'z', 'h', 'a', 'n', 'g', 'r', 'x', 'i', 'a', 'n', 'g'},
      {'2', 'z', 'h', 'a', 'n', 'g', 'r', 'x', 'i', 'a', 'n', 'g'},
      {'3', 'z', 'h', 'a', 'n', 'g', 'r', 'x', 'i', 'a', 'n', 'g'},
  };
  for (int i = 0; i < 3; ++i) {
    for (int j = 0; j < 13; ++j) {
      printf(" %c ", str[i][j]);
      if (j == 12) {
        printf("\n");
      }
    }
  }
  char *p = (char *)str;
  char *p2 = (char *)str;
  char *p3 = (char *)str;
  printf("%s\n", p);
  printf("%s\n", ++p);
  printf("%c\n", *(++p));
  printf("--------------------------\n");
  printf("%s\n", p2);
  for (int k = 0; k < 39; ++k) {
    printf(" %c ", *(p2++));
    if ((k + 1) % 13 == 0) {
      printf("\n");
    }
  }
  while (*p3 != '\0') {
    printf(" %c ", *(p3++));
  }
  printf("\n");
  char *p4 = (char *)str[1];
  printf("%s\n", p4);
}
```