#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int main() {
  ios::sync_with_stdio(false);
  int p, res = 1;
  cin >> p;
  --p;
  for (int i = 2; i * i <= p; ++i)
    if (p % i == 0) {
      res *= i - 1;
      p /= i;
      while (p % i == 0) res *= i, p /= i;
    }
  if (p - 1) res *= p - 1;
  cout << res << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

The cows have just learned what a primitive root is! Given a prime p, a primitive root <image> is an integer x (1 ≤ x < p) such that none of integers x - 1, x2 - 1, ..., xp - 2 - 1 are divisible by p, but xp - 1 - 1 is. 

Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots <image>.

Input

The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.

Output

Output on a single line the number of primitive roots <image>.

Examples

Input

3


Output

1


Input

5


Output

2

Note

The only primitive root <image> is 2.

The primitive roots <image> are 2 and 3.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int main() {
  ios::sync_with_stdio(false);
  int p, res = 1;
  cin >> p;
  --p;
  for (int i = 2; i * i <= p; ++i)
    if (p % i == 0) {
      res *= i - 1;
      p /= i;
      while (p % i == 0) res *= i, p /= i;
    }
  if (p - 1) res *= p - 1;
  cout << res << endl;
  return 0;
}

```