#include <bits/stdc++.h>
using namespace std;
int P;
bool chack(int n) {
  int i, Mod = n;
  for (i = 1; i < P - 2; i++) {
    Mod = (n * Mod) % P;
    if (Mod == 1) return 0;
  }
  Mod = (n * Mod) % P;
  if (Mod == 1) return 1;
  return 0;
}
int main(void) {
  int i, j = 0;
  cin >> P;
  if (P == 2) j = 1;
  for (i = 2; i < P; i++)
    if (chack(i)) {
      j++;
    }
  cout << j << "\n";
}

### 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 P;
bool chack(int n) {
  int i, Mod = n;
  for (i = 1; i < P - 2; i++) {
    Mod = (n * Mod) % P;
    if (Mod == 1) return 0;
  }
  Mod = (n * Mod) % P;
  if (Mod == 1) return 1;
  return 0;
}
int main(void) {
  int i, j = 0;
  cin >> P;
  if (P == 2) j = 1;
  for (i = 2; i < P; i++)
    if (chack(i)) {
      j++;
    }
  cout << j << "\n";
}

```