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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x9000(%rsp),%r11
L83:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L83
  sub    $0xc60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x9c60(%rbp),%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x9c5c(%rbp)
  jmp    L91
L48:
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x9c50(%rbp),%rdx
  mov    -0x9c5c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x9c5c(%rbp)
L91:
  mov    -0x9c60(%rbp),%eax
  cmp    %eax,-0x9c5c(%rbp)
  jl     L48
  movl   $0x0,-0x9c5c(%rbp)
  jmp    L84
L70:
  movl   $0x0,-0x9c58(%rbp)
  jmp    L24
L79:
  mov    -0x9c58(%rbp),%eax
  cltq
  mov    -0x9c50(%rbp,%rax,4),%edx
  mov    -0x9c58(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x9c50(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jle    L13
  mov    -0x9c58(%rbp),%eax
  cltq
  mov    -0x9c50(%rbp,%rax,4),%eax
  mov    %eax,-0x9c54(%rbp)
  mov    -0x9c58(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x9c50(%rbp,%rax,4),%edx
  mov    -0x9c58(%rbp),%eax
  cltq
  mov    %edx,-0x9c50(%rbp,%rax,4)
  mov    -0x9c58(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    -0x9c54(%rbp),%edx
  mov    %edx,-0x9c50(%rbp,%rax,4)
L13:
  addl   $0x1,-0x9c58(%rbp)
L24:
  mov    -0x9c60(%rbp),%eax
  sub    -0x9c5c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x9c58(%rbp)
  jl     L79
  addl   $0x1,-0x9c5c(%rbp)
L84:
  mov    -0x9c60(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x9c5c(%rbp)
  jl     L70
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x9c5c(%rbp)
  jmp    L56
L47:
  mov    -0x9c5c(%rbp),%eax
  cltq
  mov    -0x9c50(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x9c5c(%rbp)
L56:
  mov    -0x9c60(%rbp),%eax
  cmp    %eax,-0x9c5c(%rbp)
  jl     L47
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L34
  call   <__stack_chk_fail@plt>
L34:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D6
type: string
name: retrieve_data
arguments: data_label: D49
type: string
name: retrieve_data
arguments: data_label: D9
type: string
name: retrieve_data
arguments: data_label: D35
type: string
name: retrieve_data
arguments: data_label: D41
type: string

tool: D6: Enter the no. of numbers to be sorted :

tool: D49: %d

tool: D9: Enter Elements:

tool: D35: 
The sorted numbers are :

tool: D41: %d 

assistant: ```c
int main() {
  int a[10000], i, j, n, t;
  printf("Enter the no. of numbers to be sorted :");
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    printf("Enter Elements:");
    scanf("%d", &a[i]);
  }
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < n - i - 1; j++) {
      if (a[j] > a[j + 1]) {
        t = a[j];
        a[j] = a[j + 1];
        a[j + 1] = t;
      }
    }
  }
  printf("\nThe sorted numbers are :");
  for (i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  return 0;
}
```