user: 请从给定的汇编指令推导出C源代码：

```asm
<six>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movabs $0x7872676e61687a31,%rax
  mov    %rax,0x10(%rsp)
  movl   $0x676e6169,0x18(%rsp)
  movb   $0x0,0x1c(%rsp)
  lea    0x1(%rax),%rax
  mov    %rax,0x1d(%rsp)
  movl   $0x676e6169,0x25(%rsp)
  movb   $0x0,0x29(%rsp)
  lea    0x1(%rax),%rax
  mov    %rax,0x2a(%rsp)
  movl   $0x676e6169,0x32(%rsp)
  movb   $0x0,0x36(%rsp)
  lea    0x1d(%rsp),%rbp
  lea    0x44(%rsp),%r13
  lea    D43(%rip),%r12
  jmp    L76
L62:
  movsbl (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L62
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0xd,%rbp
  cmp    %r13,%rbp
  je     L22
L76:
  movsbl -0xd(%rbp),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rbx
  jmp    L62
L22:
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rdi
  call   <puts@plt>
  lea    0x11(%rsp),%rdi
  call   <puts@plt>
  movsbl 0x12(%rsp),%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D31(%rip),%rdi
  call   <puts@plt>
  mov    %rbx,%rdi
  call   <puts@plt>
  movsbl 0x10(%rsp),%esi
  lea    D43(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x11(%rsp),%rbx
  lea    D43(%rip),%r13
  mov    $0x30,%ebp
  lea    0x40(%rsp),%rax
  mov    %rax,0x8(%rsp)
  sub    0x8(%rsp),%ebp
  lea    0x37(%rsp),%r12
  jmp    L39
L56:
  cmp    %r12,%rbx
  je     L91
L39:
  add    $0x1,%rbx
  movsbl -0x1(%rbx),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x0(%rbp,%rbx,1),%edx
  movslq %edx,%rax
  imul   $0x4ec4ec4f,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,2),%ecx
  lea    (%rax,%rcx,4),%eax
  cmp    %eax,%edx
  jne    L56
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L56
L91:
  movzbl 0x10(%rsp),%esi
  test   %sil,%sil
  je     L103
  lea    0x10(%rsp),%rbx
  lea    D43(%rip),%rbp
L83:
  add    $0x1,%rbx
  movsbl %sil,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movzbl (%rbx),%esi
  test   %sil,%sil
  jne    L83
L103:
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    0x1d(%rsp),%rdi
  call   <puts@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D43
datatype: string
name: unpack_data
arguments: tag: D4
datatype: string
name: unpack_data
arguments: tag: D31
datatype: string

tool: D43:  %c 

tool: D4: %c


tool: D31: --------------------------

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);
}
```