#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
int main() {
  int p;
  cin >> p;
  int n = 0;
  for (int(x) = (1); x < (p); ++(x)) {
    long long xx = x;
    bool b = 1;
    for (int(k) = (1); k < (p - 1); ++(k)) {
      if ((xx - 1) % p == 0) {
        b = false;
        break;
      }
      xx *= x;
      xx %= p;
    }
    if (b && (xx - 1) % p == 0) n++;
  }
  cout << 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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
int main() {
  int p;
  cin >> p;
  int n = 0;
  for (int(x) = (1); x < (p); ++(x)) {
    long long xx = x;
    bool b = 1;
    for (int(k) = (1); k < (p - 1); ++(k)) {
      if ((xx - 1) % p == 0) {
        b = false;
        break;
      }
      xx *= x;
      xx %= p;
    }
    if (b && (xx - 1) % p == 0) n++;
  }
  cout << n;
  return 0;
}

```