#include <bits/stdc++.h>
using namespace std;
int main() {
  int p;
  cin >> p;
  float a = 0;
  vector<int> v(p + 1);
  for (int i = 2; i < p; i++) {
    if (v[i] == 1) {
      continue;
    }
    for (int j = 2; j * i <= p; j++) {
      v[i * j] = 1;
    }
  }
  float count = 1;
  for (int i = 2; i < p; i++) {
    if (v[i] == 0 && (p - 1) % i == 0) {
      a = i;
      count = count * (1 - (1 / a));
    }
  }
  cout << count * (p - 1);
}

### 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>
using namespace std;
int main() {
  int p;
  cin >> p;
  float a = 0;
  vector<int> v(p + 1);
  for (int i = 2; i < p; i++) {
    if (v[i] == 1) {
      continue;
    }
    for (int j = 2; j * i <= p; j++) {
      v[i * j] = 1;
    }
  }
  float count = 1;
  for (int i = 2; i < p; i++) {
    if (v[i] == 0 && (p - 1) % i == 0) {
      a = i;
      count = count * (1 - (1 / a));
    }
  }
  cout << count * (p - 1);
}

```