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

```