#include <bits/stdc++.h>
const int N = 10100;
using namespace std;
int p;
bool f(int x) {
  int y = 1;
  for (int i = 1; i <= p - 1; i++) {
    y = (y * x) % p;
    int nx = (y - 1 + p) % p;
    if (nx == 0 && i != p - 1) return false;
    if (nx != 0 && i == p - 1) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> p;
  int cnt = 0;
  for (int i = 1; i < p; i++) cnt += f(i);
  cout << cnt << endl;
}

### Prompt

Generate a cpp solution to 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>
const int N = 10100;
using namespace std;
int p;
bool f(int x) {
  int y = 1;
  for (int i = 1; i <= p - 1; i++) {
    y = (y * x) % p;
    int nx = (y - 1 + p) % p;
    if (nx == 0 && i != p - 1) return false;
    if (nx != 0 && i == p - 1) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> p;
  int cnt = 0;
  for (int i = 1; i < p; i++) cnt += f(i);
  cout << cnt << endl;
}

```