user: 请从给定的汇编指令推导出C源代码：

```asm
<six>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x7872676e61687a31,%rax
  mov    %rax,-0x30(%rbp)
  movl   $0x676e6169,-0x28(%rbp)
  movb   $0x0,-0x24(%rbp)
  movabs $0x7872676e61687a32,%rax
  mov    %rax,-0x23(%rbp)
  movl   $0x676e6169,-0x1b(%rbp)
  movb   $0x0,-0x17(%rbp)
  movabs $0x7872676e61687a33,%rax
  mov    %rax,-0x16(%rbp)
  movl   $0x676e6169,-0xe(%rbp)
  movb   $0x0,-0xa(%rbp)
  movl   $0x0,-0x5c(%rbp)
  jmp    L95
L52:
  movl   $0x0,-0x58(%rbp)
  jmp    L57
L25:
  mov    -0x58(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x5c(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rbp,%rax
  add    %rcx,%rax
  sub    $0x30,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0xc,-0x58(%rbp)
  jne    L14
  mov    $0xa,%edi
  call   <putchar@plt>
L14:
  addl   $0x1,-0x58(%rbp)
L57:
  cmpl   $0xc,-0x58(%rbp)
  jle    L25
  addl   $0x1,-0x5c(%rbp)
L95:
  cmpl   $0x2,-0x5c(%rbp)
  jle    L52
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x40(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x50(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x48(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addq   $0x1,-0x40(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addq   $0x1,-0x40(%rbp)
  mov    -0x40(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x54(%rbp)
  jmp    L92
L82:
  mov    -0x50(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x50(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x54(%rbp),%eax
  lea    0x1(%rax),%ecx
  movslq %ecx,%rax
  imul   $0x4ec4ec4f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  test   %edx,%edx
  jne    L86
  mov    $0xa,%edi
  call   <putchar@plt>
L86:
  addl   $0x1,-0x54(%rbp)
L92:
  cmpl   $0x26,-0x54(%rbp)
  jle    L82
  jmp    L27
L47:
  mov    -0x48(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x48(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L27:
  mov    -0x48(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L47
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    -0x30(%rbp),%rax
  add    $0xd,%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D42
datatype: string
name: extract_symbolic_data
arguments: data_tag: D28
datatype: string
name: extract_symbolic_data
arguments: data_tag: D30
datatype: string

tool: D42:  %c 

tool: D28: %c


tool: D30: --------------------------

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);
}
```