#include <bits/stdc++.h>
int ans, p, i;
bool f(int x) {
  int t, y, i;
  t = 1;
  for (i = 1; i < p - 1; i++) {
    t = t * x % p;
    if (t - 1 == 0) {
      return 0;
    }
  }
  t = t * x % p;
  return (t - 1 == 0);
}
int main() {
  scanf("%d\n", &p);
  ans = 0;
  for (i = 1; i < p; i++) {
    if (f(i)) {
      ans++;
    }
  }
  printf("%d\n", ans);
  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>
int ans, p, i;
bool f(int x) {
  int t, y, i;
  t = 1;
  for (i = 1; i < p - 1; i++) {
    t = t * x % p;
    if (t - 1 == 0) {
      return 0;
    }
  }
  t = t * x % p;
  return (t - 1 == 0);
}
int main() {
  scanf("%d\n", &p);
  ans = 0;
  for (i = 1; i < p; i++) {
    if (f(i)) {
      ans++;
    }
  }
  printf("%d\n", ans);
  return 0;
}

```