user: What would the C source look like for the assembly code provided here:

```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    L75
  lea    0x2c(%rsp),%rax
  movl   $0x0,0xc(%rsp)
  lea    0x30(%rsp),%r13
  mov    %rax,0x10(%rsp)
  mov    %r13,%rbx
  nopl   0x0(%rax)
L37:
  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    L9
L87:
  mov    %r14,%rsi
  mov    %r12,%rdi
  add    $0x1,%r15d
  add    $0x4,%r14
  call   <scanf@plt>
  mov    0x2c(%rsp),%eax
  cmp    %r15d,%eax
  jg     L87
  test   %eax,%eax
  jle    L9
  xor    %edx,%edx
L60:
  mov    0x0(%rbp,%rdx,4),%ecx
  mov    %edx,%esi
  cmp    %edx,%ecx
  je     L21
  lea    0x1(%rdx),%edi
  cmp    %edi,%ecx
  je     L21
  test   %edx,%edx
  jne    L0
L33:
  lea    D0(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
L9:
  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     L37
  test   %eax,%eax
  jle    L75
  xor    %ebx,%ebx
  lea    D27(%rip),%rbp
  nopl   0x0(%rax)
L123:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  add    $0x4,%r13
  call   <printf@plt>
  cmp    %ebx,0x28(%rsp)
  jg     L123
L75:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L114
  add    $0x68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  lea    -0x1(%rdx),%edi
  cmp    %edi,%ecx
  jne    L33
  nopl   0x0(%rax)
L21:
  lea    -0x1(%rax),%ecx
  cmp    %esi,%ecx
  je     L38
L54:
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L60
  jmp    L9
L38:
  lea    D13(%rip),%rsi
  mov    %rbx,%rdi
  mov    %rdx,0x18(%rsp)
  call   <strcpy@plt>
  mov    0x2c(%rsp),%eax
  mov    0x18(%rsp),%rdx
  jmp    L54
L114:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D15
type: string
name: extract_symbolic_data
arguments: data_label: D0
type: string
name: extract_symbolic_data
arguments: data_label: D27
type: string
name: extract_symbolic_data
arguments: data_label: D13
type: string

tool: D15: %d

tool: D0: No

tool: D27: %s


tool: D13: 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;
}
```