#include <bits/stdc++.h>
using namespace std;
int prim[2500];
int main() {
  int i, j, n, sum, k;
  prim[1] = 1;
  while (scanf("%d", &n) != EOF) {
    k = 0;
    for (i = 1; i < n; i++) {
      sum = 1;
      for (j = 1; j < n; j++) {
        if (j == 1)
          sum = sum * i - 1;
        else
          sum = (sum + 1) * i - 1;
        if (sum % n == 0) {
          break;
        }
        sum %= n;
      }
      if (j == n - 1) {
        k++;
      }
    }
    printf("%d\n", k);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 prim[2500];
int main() {
  int i, j, n, sum, k;
  prim[1] = 1;
  while (scanf("%d", &n) != EOF) {
    k = 0;
    for (i = 1; i < n; i++) {
      sum = 1;
      for (j = 1; j < n; j++) {
        if (j == 1)
          sum = sum * i - 1;
        else
          sum = (sum + 1) * i - 1;
        if (sum % n == 0) {
          break;
        }
        sum %= n;
      }
      if (j == n - 1) {
        k++;
      }
    }
    printf("%d\n", k);
  }
  return 0;
}

```