#include <bits/stdc++.h>
using namespace std;
int check(int p, int x) {
  int k = x;
  for (int i = 1; i < p - 1; i++) x = (x * k) % p;
  return x;
}
int main() {
  int p;
  cin >> p;
  int res = 0;
  for (int x = 1; x < p; ++x) {
    int k = check(p, x);
    if (k % p == 1) {
      int count = 0;
      for (int i = p - 2; i >= 1; i--) {
        k = (k * x) % p;
        if (k % p != 1) count++;
      }
      if (count == p - 2) res++;
    }
  }
  cout << res;
}

### 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>
using namespace std;
int check(int p, int x) {
  int k = x;
  for (int i = 1; i < p - 1; i++) x = (x * k) % p;
  return x;
}
int main() {
  int p;
  cin >> p;
  int res = 0;
  for (int x = 1; x < p; ++x) {
    int k = check(p, x);
    if (k % p == 1) {
      int count = 0;
      for (int i = p - 2; i >= 1; i--) {
        k = (k * x) % p;
        if (k % p != 1) count++;
      }
      if (count == p - 2) res++;
    }
  }
  cout << res;
}

```