#include <bits/stdc++.h>
using namespace std;
long long int i, N, k, p, j, Ans;
int main() {
  cin >> N;
  for (i = 1; i < N; i++) {
    bool ok = false;
    p = 1;
    for (j = 1; j <= N - 2; j++) {
      p *= i;
      if ((p - 1) % N == 0) {
        ok = true;
        break;
      }
      p %= N;
    }
    p *= i;
    if (ok == false && ((p - 1)) % N == 0) Ans++;
  }
  cout << Ans << 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;
long long int i, N, k, p, j, Ans;
int main() {
  cin >> N;
  for (i = 1; i < N; i++) {
    bool ok = false;
    p = 1;
    for (j = 1; j <= N - 2; j++) {
      p *= i;
      if ((p - 1) % N == 0) {
        ok = true;
        break;
      }
      p %= N;
    }
    p *= i;
    if (ok == false && ((p - 1)) % N == 0) Ans++;
  }
  cout << Ans << endl;
  return 0;
}

```