#include <bits/stdc++.h>
using namespace std;
bool fu(int a, int b, int c) {
  int ans = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      ans *= a;
      ans %= c;
    }
    a *= a;
    a %= c;
    b /= 2;
  }
  return ((ans - 1) % c == 0);
}
int main() {
  int p;
  cin >> p;
  int ans = 0;
  for (int x = 1; x < p; x++) {
    bool flag = true;
    for (int k = 1; k < p - 1; k++) {
      if (fu(x, k, p)) {
        flag = false;
        break;
      }
    }
    if (flag && (fu(x, p - 1, p))) {
      ans++;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate 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;
bool fu(int a, int b, int c) {
  int ans = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      ans *= a;
      ans %= c;
    }
    a *= a;
    a %= c;
    b /= 2;
  }
  return ((ans - 1) % c == 0);
}
int main() {
  int p;
  cin >> p;
  int ans = 0;
  for (int x = 1; x < p; x++) {
    bool flag = true;
    for (int k = 1; k < p - 1; k++) {
      if (fu(x, k, p)) {
        flag = false;
        break;
      }
    }
    if (flag && (fu(x, p - 1, p))) {
      ans++;
    }
  }
  cout << ans << endl;
  return 0;
}

```