#include <bits/stdc++.h>
const int N = 1000000000;
const int INF = 1000000000;
const long double EPS = 1E-10;
using namespace std;
int phi(int n) {
  int result = 1;
  int i = 2;
  while (i * i <= n) {
    int d = 0;
    int m = 1;
    while (n % i == 0) {
      d++;
      n /= i;
      m *= i;
    }
    if (d > 0) result *= m - m / i;
    i++;
  }
  if (n > 1) result *= n - 1;
  return result;
}
int main(int argc, char* argv[]) {
  int n;
  cin >> n;
  cout << phi(phi(n));
  return 0;
}

### Prompt

Please create a solution in cpp 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>
const int N = 1000000000;
const int INF = 1000000000;
const long double EPS = 1E-10;
using namespace std;
int phi(int n) {
  int result = 1;
  int i = 2;
  while (i * i <= n) {
    int d = 0;
    int m = 1;
    while (n % i == 0) {
      d++;
      n /= i;
      m *= i;
    }
    if (d > 0) result *= m - m / i;
    i++;
  }
  if (n > 1) result *= n - 1;
  return result;
}
int main(int argc, char* argv[]) {
  int n;
  cin >> n;
  cout << phi(phi(n));
  return 0;
}

```