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

```asm
<main>:
  endbr64
  push   %r15
  lea    D2(%rip),%r15
  push   %r14
  lea    D9(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D36(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%r13
  call   <scanf@plt>
  lea    D0(%rip),%rdi
  lea    0x38(%rsp),%r12
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L44
  nop
L136:
  mov    %r13,%rbx
  nopl   0x0(%rax,%rax,1)
L148:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbx,%r12
  jne    L148
  mov    0x14(%rsp),%eax
  cmp    %eax,0x10(%rsp)
  jg     L5
  jl     L2
L123:
  mov    %r14,%rdi
  call   <printf@plt>
L1:
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L136
L44:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L103
  add    $0x1b8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L2:
  mov    0x18(%rsp),%edx
  mov    0x1c(%rsp),%ecx
  cmp    %edx,%eax
  setge  %al
  cmp    %ecx,%edx
  setge  %dl
  movzbl %al,%eax
  movzbl %dl,%edx
  add    %edx,%eax
  mov    0x20(%rsp),%edx
  cmp    %edx,%ecx
  setge  %cl
  movzbl %cl,%ecx
  add    %ecx,%eax
  mov    0x24(%rsp),%ecx
  cmp    %ecx,%edx
  setge  %dl
  movzbl %dl,%edx
  add    %edx,%eax
  mov    0x28(%rsp),%edx
  cmp    %edx,%ecx
  setge  %cl
  movzbl %cl,%ecx
  add    %ecx,%eax
  mov    0x2c(%rsp),%ecx
  cmp    %ecx,%edx
  jge    L123
  mov    0x30(%rsp),%edx
  cmp    %ecx,%edx
  jle    L123
  cmp    0x34(%rsp),%edx
  jge    L123
L38:
  test   %eax,%eax
  jne    L123
  mov    %r15,%rdi
  call   <printf@plt>
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L5:
  mov    0x18(%rsp),%edx
  mov    0x1c(%rsp),%ecx
  cmp    %edx,%eax
  setle  %al
  cmp    %ecx,%edx
  setle  %dl
  movzbl %al,%eax
  movzbl %dl,%edx
  add    %edx,%eax
  mov    0x20(%rsp),%edx
  cmp    %edx,%ecx
  setle  %cl
  movzbl %cl,%ecx
  add    %ecx,%eax
  mov    0x24(%rsp),%ecx
  cmp    %ecx,%edx
  setle  %dl
  movzbl %dl,%edx
  add    %edx,%eax
  mov    0x28(%rsp),%edx
  cmp    %edx,%ecx
  setle  %cl
  movzbl %cl,%ecx
  add    %ecx,%eax
  mov    0x2c(%rsp),%ecx
  cmp    %ecx,%edx
  jle    L123
  mov    0x30(%rsp),%edx
  cmp    %edx,%ecx
  jle    L123
  cmp    %edx,0x34(%rsp)
  jl     L38
  jmp    L123
L103:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
datatype: string
name: retrieve_data
arguments: data_tag: D9
datatype: string
name: retrieve_data
arguments: data_tag: D36
datatype: string
name: retrieve_data
arguments: data_tag: D0
datatype: string

tool: D2: Ordered


tool: D9: Unordered


tool: D36: %d

tool: D0: Lumberjacks:


assistant: ```c
int main() {
  int a[101], i, t, x;
  scanf("%d", &t);
  printf("Lumberjacks:\n");
  while (t--) {
    x = 0;
    for (i = 0; i < 10; i++) {
      scanf("%d", &a[i]);
    }
    if (a[0] > a[1]) {
      for (i = 1; i < 10; i++) {
        if (a[i - 1] > a[i]) {
          x = x;
        } else {
          x++;
        }
      }
      if (x == 0) {
        printf("Ordered\n");
      } else {
        printf("Unordered\n");
      }
    } else if (a[0] < a[1]) {
      for (i = 1; i < 10; i++) {
        if (a[i - 1] < a[i]) {
          x = x;
        } else {
          x++;
        }
      }
      if (x == 0) {
        printf("Ordered\n");
      } else {
        printf("Unordered\n");
      }
    } else {
      printf("Unordered\n");
    }
  }
  return 0;
}
```