#include <bits/stdc++.h>
int calc(int n) {
  int res = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      res = res / i * (i - 1);
      while (n % i == 0) n /= i;
    }
  }
  if (n != 1) res = res / n * (n - 1);
  return res;
}
int main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", calc(n - 1));
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
int calc(int n) {
  int res = n;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      res = res / i * (i - 1);
      while (n % i == 0) n /= i;
    }
  }
  if (n != 1) res = res / n * (n - 1);
  return res;
}
int main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", calc(n - 1));
  return 0;
}

```