#include <bits/stdc++.h>
using namespace std;
void AWAD() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  AWAD();
  int n, i;
  cin >> n;
  int ans = 0;
  for (i = 1; i < n; i++) {
    if (gcd(n - 1, i) == 1) ++ans;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
void AWAD() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  AWAD();
  int n, i;
  cin >> n;
  int ans = 0;
  for (i = 1; i < n; i++) {
    if (gcd(n - 1, i) == 1) ++ans;
  }
  cout << ans << endl;
  return 0;
}

```