#include <bits/stdc++.h>
using namespace std;
int fastpow(int base, int p, int mod) {
  if (p == 1) return base % mod;
  int x = fastpow(base, p / 2, mod) % mod;
  if (p % 2 == 0) {
    return (x * x) % mod;
  } else {
    return (((x * x) % mod) * base % mod) % mod;
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  int counter = 0;
  for (int i = 1; i < n; i++) {
    bool c = 1;
    for (int j = 1; j < n; j++) {
      int x = (fastpow(i, j, n) - 1) % n;
      if ((x == 0 && j != n - 1)) {
        c = 0;
        break;
      }
      if (x != 0 && j == n - 1) {
        c = 0;
        break;
      }
    }
    if (c) {
      counter++;
    }
  }
  cout << counter;
  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;
int fastpow(int base, int p, int mod) {
  if (p == 1) return base % mod;
  int x = fastpow(base, p / 2, mod) % mod;
  if (p % 2 == 0) {
    return (x * x) % mod;
  } else {
    return (((x * x) % mod) * base % mod) % mod;
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  int counter = 0;
  for (int i = 1; i < n; i++) {
    bool c = 1;
    for (int j = 1; j < n; j++) {
      int x = (fastpow(i, j, n) - 1) % n;
      if ((x == 0 && j != n - 1)) {
        c = 0;
        break;
      }
      if (x != 0 && j == n - 1) {
        c = 0;
        break;
      }
    }
    if (c) {
      counter++;
    }
  }
  cout << counter;
  return 0;
}

```