#include <bits/stdc++.h>
using namespace std;
int n, s, tem, c, ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    tem = 1;
    c = 0;
    for (int j = 1; j <= n - 2; j++) {
      tem = 1ll * tem * i % n;
      if ((tem - 1 + n) % n != 0) c++;
    }
    tem = 1ll * tem * i % n;
    if ((tem - 1 + n) % n == 0) c++;
    if (c == n - 1) ans++;
  }
  printf("%d\n", ans);
  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;
int n, s, tem, c, ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    tem = 1;
    c = 0;
    for (int j = 1; j <= n - 2; j++) {
      tem = 1ll * tem * i % n;
      if ((tem - 1 + n) % n != 0) c++;
    }
    tem = 1ll * tem * i % n;
    if ((tem - 1 + n) % n == 0) c++;
    if (c == n - 1) ans++;
  }
  printf("%d\n", ans);
  return 0;
}

```