#include <bits/stdc++.h>
using namespace std;
bool isit(long long x, long long n, long long p) {
  long long t = x;
  for (long long i = 2; i <= n - 1; i++) {
    x = ((x % p) * (t % p)) % p;
    if ((x - 1) % p == 0) return false;
  }
  x *= t;
  x--;
  if (x % p == 0) return true;
}
int main() {
  long long n;
  cin >> n;
  if (n == 2) {
    cout << 1 << endl;
    return 0;
  }
  long long cnt = 0;
  for (long long i = 2; i < n; i++) {
    if (isit(i, n - 1, n)) {
      cnt++;
    }
  }
  cout << cnt << endl;
  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;
bool isit(long long x, long long n, long long p) {
  long long t = x;
  for (long long i = 2; i <= n - 1; i++) {
    x = ((x % p) * (t % p)) % p;
    if ((x - 1) % p == 0) return false;
  }
  x *= t;
  x--;
  if (x % p == 0) return true;
}
int main() {
  long long n;
  cin >> n;
  if (n == 2) {
    cout << 1 << endl;
    return 0;
  }
  long long cnt = 0;
  for (long long i = 2; i < n; i++) {
    if (isit(i, n - 1, n)) {
      cnt++;
    }
  }
  cout << cnt << endl;
  return 0;
}

```