#include <bits/stdc++.h>
int main() {
  int p;
  while (scanf("%d", &p) == 1) {
    int ans = 0;
    for (int i = 1; i < p; i++) {
      bool IS = true;
      int S = 1;
      for (int j = 1; j <= p - 1; j++) {
        S *= i, S %= p;
        if (j == p - 1) {
          if (S != 1) {
            IS = false;
            break;
          }
        } else if (S == 1) {
          IS = false;
          break;
        }
      }
      if (IS) ans++;
    }
    printf("%d\n", ans);
  }
  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>
int main() {
  int p;
  while (scanf("%d", &p) == 1) {
    int ans = 0;
    for (int i = 1; i < p; i++) {
      bool IS = true;
      int S = 1;
      for (int j = 1; j <= p - 1; j++) {
        S *= i, S %= p;
        if (j == p - 1) {
          if (S != 1) {
            IS = false;
            break;
          }
        } else if (S == 1) {
          IS = false;
          break;
        }
      }
      if (IS) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}

```