user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    D13(%rip),%rbx
L95:
  lea    -0x3c(%rbp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    -0x3c(%rbp),%esi
  lea    -0x3(%rsi),%eax
  cmp    $0x6,%eax
  ja     L95
  movslq %esi,%rax
  mov    %rax,-0x50(%rbp)
  imul   %rax,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L59:
  cmp    %rdx,%rsp
  je     L47
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L59
L47:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L112
  orq    $0x0,-0x8(%rsp,%rax,1)
L112:
  mov    %rsp,-0x48(%rbp)
  mov    %esi,%edx
  lea    D42(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%r15d
  mov    $0x0,%r14d
  lea    D28(%rip),%r13
  cmpl   $0x0,-0x3c(%rbp)
  jg     L91
L45:
  lea    D65(%rip),%rdi
  call   <puts@plt>
  mov    -0x3c(%rbp),%esi
  test   %esi,%esi
  jle    L80
  mov    -0x50(%rbp),%r15
  shl    $0x2,%r15
  mov    -0x48(%rbp),%r12
  mov    $0x0,%r14d
  lea    D15(%rip),%r13
  jmp    L4
L23:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%r12
  cmp    %ebx,-0x3c(%rbp)
  jg     L23
L128:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r14d
  add    -0x50(%rbp),%r15
  cmp    %r14d,-0x3c(%rbp)
  jle    L45
L91:
  mov    -0x48(%rbp),%rax
  lea    (%rax,%r15,4),%r12
  mov    $0x0,%ebx
  cmpl   $0x0,-0x3c(%rbp)
  jg     L23
  jmp    L128
L141:
  mov    (%r12,%rbx,4),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,-0x3c(%rbp)
  jg     L141
L64:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r14d
  mov    -0x3c(%rbp),%esi
  add    %r15,%r12
  cmp    %r14d,%esi
  jle    L80
L4:
  mov    $0x0,%ebx
  cmpl   $0x0,-0x3c(%rbp)
  jg     L141
  jmp    L64
L80:
  mov    -0x48(%rbp),%rdi
  call   <isMagic@plt>
  test   %eax,%eax
  jne    L20
  lea    D2(%rip),%rdi
  call   <puts@plt>
L114:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L20:
  lea    D14(%rip),%rdi
  call   <puts@plt>
  jmp    L114
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D13
type_guessed: string
name: retrieve_data
arguments: tag: D42
type_guessed: string
name: retrieve_data
arguments: tag: D28
type_guessed: string
name: retrieve_data
arguments: tag: D65
type_guessed: string
name: retrieve_data
arguments: tag: D15
type_guessed: string
name: retrieve_data
arguments: tag: D2
type_guessed: string
name: retrieve_data
arguments: tag: D14
type_guessed: string

tool: D13: %d

tool: D42: Farewell!Now enter the values of you own %d by %d square:


tool: D28: %d%*c

tool: D65: Your "magic square" should look like this:

tool: D15: 	%d 

tool: D2: Your square indeed is magic!

tool: D14: Sorry but your square isn't magic!

assistant: ```c
int main() {
  int n, i = 0, j;
  do
    scanf("%d", &n);
  while (n <= 2 || n >= 10);
  int square[n][n];
  printf("Farewell!Now enter the values of you own %d by %d square:\n", n, n);
  for (; i < n; i++) {
    for (j = 0; j < n; j++) {
      scanf("%d%*c", &square[i][j]);
    }
    printf("\n");
  }
  printf("Your \"magic square\" should look like this:\n");
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++)
      printf("\t%d ", square[i][j]);
    printf("\n");
  }
  if (!isMagic((int **)square, n))
    printf("Your square indeed is magic!\n");
  else
    printf("Sorry but your square isn't magic!\n");
  return 0;
}
```