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

```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
  mov    %rsp,%rax
  mov    %rax,%rbx
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x50(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x50(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x48(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L58:
  cmp    %rdx,%rsp
  je     L4
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L58
L4:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L48
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L48:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x40(%rbp)
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4c(%rbp)
  jmp    L27
L50:
  mov    -0x4c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x4c(%rbp)
L27:
  mov    -0x50(%rbp),%eax
  cmp    %eax,-0x4c(%rbp)
  jl     L50
  mov    -0x50(%rbp),%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <counting_sort@plt>
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4c(%rbp)
  jmp    L23
L1:
  mov    -0x40(%rbp),%rax
  mov    -0x4c(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  mov    %eax,%esi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4c(%rbp)
L23:
  mov    -0x50(%rbp),%eax
  cmp    %eax,-0x4c(%rbp)
  jl     L1
  mov    $0x0,%eax
  mov    %rbx,%rsp
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D33
datatype: string
name: disassemble_data
arguments: tag: D21
datatype: string
name: disassemble_data
arguments: tag: D15
datatype: string
name: disassemble_data
arguments: tag: D30
datatype: string
name: disassemble_data
arguments: tag: D49
datatype: string

tool: D33: Enter number of elements in your array: 

tool: D21: %d

tool: D15: Enter your array: 

tool: D30: Sorted Array : 

tool: D49: %d 

assistant: ```c
int main() {
  int n, i;
  printf("Enter number of elements in your array: ");
  scanf("%d", &n);
  int input[n];
  printf("Enter your array: ");
  for (i = 0; i < n; i++)
    scanf("%d", &input[i]);
  counting_sort(input, n);
  printf("Sorted Array : ");
  for (i = 0; i < n; i++)
    printf("%d ", input[i]);
  return 0;
}
```