#include <bits/stdc++.h>
using namespace std;
bool test(int x, int p) {
  int num = x;
  for (int i = 1; i < p - 1; i++) {
    if ((x - 1) % p == 0) return false;
    x = (x * num) % p;
  }
  return ((x - 1) % p) == 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int p;
  cin >> p;
  if (p == 2) {
    cout << 1 << '\n';
    return 0;
  }
  int ans = 0;
  for (int i = 2; i < p; i++) {
    if (test(i, p)) ans++;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Create a solution in CPP for 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 test(int x, int p) {
  int num = x;
  for (int i = 1; i < p - 1; i++) {
    if ((x - 1) % p == 0) return false;
    x = (x * num) % p;
  }
  return ((x - 1) % p) == 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int p;
  cin >> p;
  if (p == 2) {
    cout << 1 << '\n';
    return 0;
  }
  int ans = 0;
  for (int i = 2; i < p; i++) {
    if (test(i, p)) ans++;
  }
  cout << ans << '\n';
  return 0;
}

```