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

### Prompt

Create a solution in Cpp for 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() {
  int n, sum = 0, j, i;
  cin >> n;
  for (i = 1; i < n; i++) {
    int flag = 0;
    int x = 1;
    for (j = 1; j < n - 1; j++) {
      x = (x * i) % n;
      if (x == 1) {
        flag = 1;
        break;
      }
    }
    if (flag == 0) {
      if ((x * i - 1) % n == 0) {
        sum++;
      }
    }
  }
  cout << sum;
  return 0;
}

```