#include <bits/stdc++.h>
using namespace std;
int bigMod(int a, int p, int m) {
  if (p == 0) return 1;
  if (p % 2) {
    return ((a % m) * (bigMod(a, p - 1, m))) % m;
  } else {
    int c = bigMod(a, p / 2, m);
    return ((c % m) * (c % m)) % m;
  }
}
int main() {
  int n, i, j, k;
  long long temp, sum, m;
  while (cin >> n) {
    int ans = 0, flag;
    for (i = 1; i < n; i++) {
      flag = 0;
      for (j = 1; j < n - 1; j++) {
        if (bigMod(i, j, n) == 1) {
          flag = 1;
        }
      }
      if (flag == 0) {
        if (bigMod(i, n - 1, n) != 0) ans++;
      }
      flag = 0;
    }
    cout << ans << endl;
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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 bigMod(int a, int p, int m) {
  if (p == 0) return 1;
  if (p % 2) {
    return ((a % m) * (bigMod(a, p - 1, m))) % m;
  } else {
    int c = bigMod(a, p / 2, m);
    return ((c % m) * (c % m)) % m;
  }
}
int main() {
  int n, i, j, k;
  long long temp, sum, m;
  while (cin >> n) {
    int ans = 0, flag;
    for (i = 1; i < n; i++) {
      flag = 0;
      for (j = 1; j < n - 1; j++) {
        if (bigMod(i, j, n) == 1) {
          flag = 1;
        }
      }
      if (flag == 0) {
        if (bigMod(i, n - 1, n) != 0) ans++;
      }
      flag = 0;
    }
    cout << ans << endl;
  }
}

```