#include <bits/stdc++.h>
#pragma GCC diagnostic ignored "-Wunused-result"
using namespace std;
int main() {
  int p, ans = 0;
  scanf("%d", &p);
  for (int x = 1; x < p; x++) {
    int i, y = 1;
    for (i = 1; i <= p - 2; i++) {
      y = y * x % p;
      if (y == 1) break;
    }
    if (i == p - 1 && y * x % p == 1) ans++;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Develop a solution in CPP 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>
#pragma GCC diagnostic ignored "-Wunused-result"
using namespace std;
int main() {
  int p, ans = 0;
  scanf("%d", &p);
  for (int x = 1; x < p; x++) {
    int i, y = 1;
    for (i = 1; i <= p - 2; i++) {
      y = y * x % p;
      if (y == 1) break;
    }
    if (i == p - 1 && y * x % p == 1) ans++;
  }
  printf("%d\n", ans);
  return 0;
}

```