#include <bits/stdc++.h>
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  long long int p, i, ans = 0;
  scanf("%I64d", &p);
  for (i = 1; i <= p - 1; i++) {
    if (gcd(p - 1, i) == 1) ans++;
  }
  printf("%I64d", ans);
  return 0;
}

### Prompt

Generate a CPP solution to 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>
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  long long int p, i, ans = 0;
  scanf("%I64d", &p);
  for (i = 1; i <= p - 1; i++) {
    if (gcd(p - 1, i) == 1) ans++;
  }
  printf("%I64d", ans);
  return 0;
}

```