#include <bits/stdc++.h>
using namespace std;
long long d(int x, int i) {
  long long sum = 1;
  for (int j = 0; j < i; j++) {
    sum = sum * x;
  }
  return sum;
}
int main() {
  int p;
  cin >> p;
  int sum = 0;
  int x = 2;
  int i, j;
  for (int x = 1; x < p; x++) {
    int y = 0;
    long long S = x;
    for (int i = 1; i < p - 1; i++) {
      S = S % p;
      if ((S - 1) % p == 0) {
        y++;
        break;
      }
      S = S * x;
    }
    if (y == 0) {
      if ((S - 1) % p == 0) sum++;
    }
  }
  cout << sum;
}

### 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;
long long d(int x, int i) {
  long long sum = 1;
  for (int j = 0; j < i; j++) {
    sum = sum * x;
  }
  return sum;
}
int main() {
  int p;
  cin >> p;
  int sum = 0;
  int x = 2;
  int i, j;
  for (int x = 1; x < p; x++) {
    int y = 0;
    long long S = x;
    for (int i = 1; i < p - 1; i++) {
      S = S % p;
      if ((S - 1) % p == 0) {
        y++;
        break;
      }
      S = S * x;
    }
    if (y == 0) {
      if ((S - 1) % p == 0) sum++;
    }
  }
  cout << sum;
}

```