#include <bits/stdc++.h>
using namespace std;
int N;
int main() {
  long long i, j;
  long long second, t, r;
  second = t = r = 0;
  scanf(" %d", &N);
  if (N == 2) {
    cout << 1 << endl;
    return 0;
  }
  for (i = 2; i <= (int)N - 1; i++) {
    second = i;
    for (j = 1; j <= (int)N; j++) {
      second %= N;
      if ((second - 1) % N == 0) break;
      second = second * i;
    }
    if (j == N - 1) t++;
  }
  cout << t << 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;
int N;
int main() {
  long long i, j;
  long long second, t, r;
  second = t = r = 0;
  scanf(" %d", &N);
  if (N == 2) {
    cout << 1 << endl;
    return 0;
  }
  for (i = 2; i <= (int)N - 1; i++) {
    second = i;
    for (j = 1; j <= (int)N; j++) {
      second %= N;
      if ((second - 1) % N == 0) break;
      second = second * i;
    }
    if (j == N - 1) t++;
  }
  cout << t << endl;
  return 0;
}

```