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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L26
L48:
  movl   $0x0,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L42
L5:
  mov    -0x18(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    -0x18(%rbp),%esi
  movslq %esi,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %esi,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %esi,%edx
  imul   %edx,%ecx
  mov    %ecx,%esi
  mov    -0x18(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %esi,%eax
  imul   %edx,%eax
  add    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x18(%rbp)
L42:
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jne    L5
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jne    L19
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L34
L19:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L34:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x18(%rbp)
L26:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  jle    L48
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D34
type: string
name: read_data
arguments: data_tag: D6
type: string
name: read_data
arguments: data_tag: D14
type: string
name: read_data
arguments: data_tag: D15
type: string

tool: D34: Enter a two numbers (n,m) m > n:

tool: D6: %d %d

tool: D14: %3d is an Armstrong number.


tool: D15: %3d is not an Armstrong number.


assistant: ```c
int main() {
  int n, m, res, nBack;
  printf("Enter a two numbers (n,m) m > n:\n");
  scanf("%d %d", &n, &m);
  for (n; n <= m; n++) {
    res = 0;
    nBack = n;
    while (n != 0) {
      res += ((n % 10) * (n % 10) * (n % 10));
      n /= 10;
    }
    if (nBack == res)
      printf("%3d is an Armstrong number.\n", nBack);
    else
      printf("%3d is not an Armstrong number.\n", nBack);
    n = nBack;
  }
  return 0;
}
```