#include <bits/stdc++.h>
using namespace std;
int primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47};
int phi(int n) {
  int num = 1, root = (int)sqrt(n), *prime = primes, Phi = n, N = n;
  for (; *prime <= root && n != 1; prime++)
    if (!(n % *prime)) {
      Phi = Phi / (*prime) * (*prime - 1);
      for (; !(n % *prime); n /= *prime, num *= *prime)
        ;
    }
  if (num != N) {
    num = N / num;
    Phi = Phi / num * (num - 1);
  }
  return Phi;
}
int main() {
  int p;
  while (scanf("%d", &p) != EOF) printf("%d\n", phi(phi(p)));
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
int primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47};
int phi(int n) {
  int num = 1, root = (int)sqrt(n), *prime = primes, Phi = n, N = n;
  for (; *prime <= root && n != 1; prime++)
    if (!(n % *prime)) {
      Phi = Phi / (*prime) * (*prime - 1);
      for (; !(n % *prime); n /= *prime, num *= *prime)
        ;
    }
  if (num != N) {
    num = N / num;
    Phi = Phi / num * (num - 1);
  }
  return Phi;
}
int main() {
  int p;
  while (scanf("%d", &p) != EOF) printf("%d\n", phi(phi(p)));
  return 0;
}

```