user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    D15(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0x28(%rsp),%rsi
  call   <scanf@plt>
  mov    0x28(%rsp),%edx
  test   %edx,%edx
  jle    L31
  lea    0x2c(%rsp),%rax
  movl   $0x0,0xc(%rsp)
  lea    0x30(%rsp),%r13
  mov    %rax,0x10(%rsp)
  mov    %r13,%rbx
  nopl   0x0(%rax)
L121:
  mov    $0x4,%edx
  xor    %esi,%esi
  mov    %rbx,%rdi
  xor    %r15d,%r15d
  call   <memset@plt>
  mov    0x10(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x2c(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rax,%r14
  mov    0x2c(%rsp),%eax
  test   %eax,%eax
  jle    L55
L99:
  mov    %r14,%rsi
  mov    %r12,%rdi
  add    $0x1,%r15d
  add    $0x4,%r14
  call   <scanf@plt>
  mov    0x2c(%rsp),%eax
  cmp    %r15d,%eax
  jg     L99
  test   %eax,%eax
  jle    L55
  xor    %edx,%edx
L33:
  mov    0x0(%rbp,%rdx,4),%ecx
  mov    %edx,%esi
  cmp    %edx,%ecx
  je     L61
  lea    0x1(%rdx),%edi
  cmp    %edi,%ecx
  je     L61
  test   %edx,%edx
  jne    L100
L7:
  lea    D24(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
L55:
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <free@plt>
  addl   $0x1,0xc(%rsp)
  mov    0x28(%rsp),%eax
  mov    0xc(%rsp),%ecx
  cmp    %ecx,%eax
  jg     L121
  test   %eax,%eax
  jle    L31
  xor    %ebx,%ebx
  lea    D2(%rip),%rbp
  nopl   0x0(%rax)
L40:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  add    $0x4,%r13
  call   <printf@plt>
  cmp    %ebx,0x28(%rsp)
  jg     L40
L31:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L100:
  lea    -0x1(%rdx),%edi
  cmp    %edi,%ecx
  jne    L7
  nopl   0x0(%rax)
L61:
  lea    -0x1(%rax),%ecx
  cmp    %esi,%ecx
  je     L150
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L33
  jmp    L55
L150:
  lea    D21(%rip),%rsi
  mov    %rbx,%rdi
  mov    %rdx,0x18(%rsp)
  call   <strcpy@plt>
  mov    0x18(%rsp),%rdx
  mov    0x2c(%rsp),%eax
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L33
  jmp    L55
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D15
datatype: string
name: retrieve_data
arguments: tag: D24
datatype: string
name: retrieve_data
arguments: tag: D2
datatype: string
name: retrieve_data
arguments: tag: D21
datatype: string

tool: D15: %d

tool: D24: No

tool: D2: %s


tool: D21: Yes

assistant: ```c
int main() {
  int q;
  scanf("%d", &q);
  char output[10][4];
  for (int a0 = 0; a0 < q; a0++) {
    memset(output[a0], 0, 4);
    int n;
    scanf("%d", &n);
    int *a = malloc(sizeof(int) * n);
    for (int a_i = 0; a_i < n; a_i++) {
      scanf("%d", &a[a_i]);
    }
    for (int a_i = 0; a_i < n; a_i++) {
      if (a[a_i] == a_i || (a[a_i] == (a_i + 1) || (0 <= (a_i - 1) && a[a_i] == (a_i - 1)))) {
        if (a_i == n - 1) {
          strcpy(output[a0], "Yes");
        }
      } else {
        strcpy(output[a0], "No");
        break;
      }
    }
    free(a);
  }
  for (int a0 = 0; a0 < q; a0++) {
    printf("%s\n", output[a0]);
  }
  return 0;
}
```