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

```asm
<primitive_root>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %edi,-0xd4(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xcc(%rbp)
  mov    -0xd4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc8(%rbp)
L54:
  movl   $0x2,-0xc4(%rbp)
  jmp    L112
L99:
  mov    -0xc8(%rbp),%eax
  mov    $0x0,%edx
  divl   -0xc4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L105
  mov    -0xc8(%rbp),%eax
  mov    $0x0,%edx
  divl   -0xc4(%rbp)
  mov    %eax,-0xc8(%rbp)
  mov    -0xcc(%rbp),%eax
  mov    -0xc4(%rbp),%edx
  mov    %edx,-0xb0(%rbp,%rax,4)
  cmpl   $0x0,-0xcc(%rbp)
  jne    L43
  addl   $0x1,-0xcc(%rbp)
  jmp    L68
L43:
  mov    -0xcc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%eax
  mov    -0xb0(%rbp,%rax,4),%eax
  cmp    %eax,-0xc4(%rbp)
  je     L68
  addl   $0x1,-0xcc(%rbp)
  jmp    L68
L105:
  addl   $0x1,-0xc4(%rbp)
L112:
  mov    -0xc4(%rbp),%eax
  cmp    -0xc8(%rbp),%eax
  jbe    L99
  jmp    L62
L68:
  nop
L62:
  cmpl   $0x1,-0xc8(%rbp)
  jbe    L74
  cmpl   $0x27,-0xcc(%rbp)
  jbe    L54
L74:
  movl   $0x0,-0xbc(%rbp)
  movl   $0x2,-0xc0(%rbp)
  jmp    L101
L157:
  movl   $0x1,-0xb8(%rbp)
  movl   $0x0,-0xc4(%rbp)
  jmp    L58
L52:
  mov    -0xd4(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0xc4(%rbp),%eax
  mov    -0xb0(%rbp,%rax,4),%esi
  mov    %ecx,%eax
  mov    $0x0,%edx
  div    %esi
  mov    %eax,-0xb4(%rbp)
  mov    -0xd4(%rbp),%edx
  mov    -0xb4(%rbp),%ecx
  mov    -0xc0(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <modpow@plt>
  cmp    $0x1,%eax
  jne    L40
  movl   $0x0,-0xb8(%rbp)
  jmp    L142
L40:
  addl   $0x1,-0xc4(%rbp)
L58:
  mov    -0xc4(%rbp),%eax
  cmp    -0xcc(%rbp),%eax
  jb     L52
L142:
  cmpl   $0x0,-0xb8(%rbp)
  je     L14
  movl   $0x1,-0xbc(%rbp)
  jmp    L185
L14:
  addl   $0x1,-0xc0(%rbp)
L101:
  mov    -0xc0(%rbp),%eax
  cmp    -0xd4(%rbp),%eax
  jb     L157
L185:
  cmpl   $0x0,-0xbc(%rbp)
  jne    L123
  mov    -0xd4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L123:
  mov    -0xc0(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L144
  call   <__stack_chk_fail@plt>
L144:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_infered: string

tool: D7: warning: primitive_root(), could not find primitive root of %u


assistant: ```c
unsigned int primitive_root(unsigned int _n) {
  unsigned int unique_factors[(40)];
  unsigned int num_unique_factors = 0;
  unsigned int n = _n - 1;
  unsigned int k;
  do {
    for (k = 2; k <= n; k++) {
      if ((n % k) == 0) {
        n /= k;
        unique_factors[num_unique_factors] = k;
        if (num_unique_factors == 0)
          num_unique_factors++;
        else if (unique_factors[num_unique_factors - 1] != k)
          num_unique_factors++;
        break;
      }
    }
  } while (n > 1 && num_unique_factors < (40));
  unsigned int g;
  int root_found = 0;
  for (g = 2; g < _n; g++) {
    int is_root = 1;
    for (k = 0; k < num_unique_factors; k++) {
      unsigned int e = (_n - 1) / unique_factors[k];
      if (modpow(g, e, _n) == 1) {
        is_root = 0;
        break;
      }
    }
    if (is_root) {
      root_found = 1;
      break;
    }
  }
  if (!root_found)
    printf("warning: primitive_root(), could not find primitive root of %u\n", _n);
  return g;
}
```