#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long p;
  cin >> p;
  long long ans = 0;
  for (long long x = 1; x <= p; ++x) {
    long long t = x;
    long long j = 1;
    bool f = 1;
    while (j <= p - 2) {
      if (t % p == 1) f = 0;
      j++;
      t = (t * x) % p;
    }
    if ((t - 1) % p) f = 0;
    if (f) ans++;
  }
  cout << ans;
  return 0;
}

### Prompt

Your task is to create 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;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long p;
  cin >> p;
  long long ans = 0;
  for (long long x = 1; x <= p; ++x) {
    long long t = x;
    long long j = 1;
    bool f = 1;
    while (j <= p - 2) {
      if (t % p == 1) f = 0;
      j++;
      t = (t * x) % p;
    }
    if ((t - 1) % p) f = 0;
    if (f) ans++;
  }
  cout << ans;
  return 0;
}

```