#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
  if (b == 0) {
    return a;
  }
  if (a == 1 || b == 1) {
    return 1;
  }
  return gcd(b, a % b);
}
int main() {
  int a, b, count = 0;
  cin >> a;
  for (b = 1; b < a - 1; ++b) {
    if (gcd(a - 1, b) == 1) {
      ++count;
    }
  }
  if (a == 2) {
    cout << 1 << endl;
  } else {
    cout << count << endl;
  }
  return 0;
}

### Prompt

Generate 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 gcd(int a, int b) {
  if (b == 0) {
    return a;
  }
  if (a == 1 || b == 1) {
    return 1;
  }
  return gcd(b, a % b);
}
int main() {
  int a, b, count = 0;
  cin >> a;
  for (b = 1; b < a - 1; ++b) {
    if (gcd(a - 1, b) == 1) {
      ++count;
    }
  }
  if (a == 2) {
    cout << 1 << endl;
  } else {
    cout << count << endl;
  }
  return 0;
}

```