#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  if (n == 2) {
    cout << 1;
    return 0;
  }
  int c = 0;
  for (int i = 2; i < n; ++i) {
    bool f = true;
    int arr[2005] = {0};
    for (long long j = i, s = 1; s <= n - 2; j = (j * i) % n, s++) {
      if (arr[j] != 0) {
        f = false;
        break;
      }
      arr[j] = j;
    }
    cout << endl;
    if (f) c++;
  }
  cout << c;
  return 0;
}

### Prompt

Please formulate a CPP solution 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() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  if (n == 2) {
    cout << 1;
    return 0;
  }
  int c = 0;
  for (int i = 2; i < n; ++i) {
    bool f = true;
    int arr[2005] = {0};
    for (long long j = i, s = 1; s <= n - 2; j = (j * i) % n, s++) {
      if (arr[j] != 0) {
        f = false;
        break;
      }
      arr[j] = j;
    }
    cout << endl;
    if (f) c++;
  }
  cout << c;
  return 0;
}

```