#include <bits/stdc++.h>
using namespace std;
void help() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  help();
  int p, cnt = 0;
  cin >> p;
  if (p == 2)
    cout << 1 << endl;
  else {
    for (int i = 2; i < p; i++) {
      int x = i;
      for (int j = 2; j < p - 1; j++) {
        if ((x - 1) % p == 0) break;
        x *= i;
        x %= p;
      }
      if ((x - 1) % p != 0) cnt++;
    }
    cout << cnt << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 help() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  help();
  int p, cnt = 0;
  cin >> p;
  if (p == 2)
    cout << 1 << endl;
  else {
    for (int i = 2; i < p; i++) {
      int x = i;
      for (int j = 2; j < p - 1; j++) {
        if ((x - 1) % p == 0) break;
        x *= i;
        x %= p;
      }
      if ((x - 1) % p != 0) cnt++;
    }
    cout << cnt << endl;
  }
  return 0;
}

```