#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  bool f;
  while (scanf("%d", &n) != EOF) {
    int c = 0;
    for (int i = 1; i < n; i++) {
      f = true;
      int s = i % n;
      for (int j = 1; j < n - 1; j++) {
        if (s == 1) {
          f = false;
          break;
        }
        s = (s * i) % n;
      }
      if (f && s % n == 1) c++;
    }
    cout << c << endl;
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
using namespace std;
int main() {
  int n;
  bool f;
  while (scanf("%d", &n) != EOF) {
    int c = 0;
    for (int i = 1; i < n; i++) {
      f = true;
      int s = i % n;
      for (int j = 1; j < n - 1; j++) {
        if (s == 1) {
          f = false;
          break;
        }
        s = (s * i) % n;
      }
      if (f && s % n == 1) c++;
    }
    cout << c << endl;
  }
  return 0;
}

```