#include <bits/stdc++.h>
using namespace std;
const int inf = ~0u >> 2;
const int MOD = 1e7 + 7;
const int N = 1e5 + 10;
int main() {
  long long p, tmp, ct = 0;
  bool flag;
  cin >> p;
  for (int i = 1; i <= p - 1; ++i) {
    tmp = 1;
    flag = 1;
    for (int j = 1; j <= p - 2; ++j) {
      tmp = tmp * i % p;
      if ((tmp - 1) % p == 0) flag = 0;
    }
    if (flag && ((tmp * i - 1) % p == 0)) ++ct;
  }
  cout << ct << endl;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
const int inf = ~0u >> 2;
const int MOD = 1e7 + 7;
const int N = 1e5 + 10;
int main() {
  long long p, tmp, ct = 0;
  bool flag;
  cin >> p;
  for (int i = 1; i <= p - 1; ++i) {
    tmp = 1;
    flag = 1;
    for (int j = 1; j <= p - 2; ++j) {
      tmp = tmp * i % p;
      if ((tmp - 1) % p == 0) flag = 0;
    }
    if (flag && ((tmp * i - 1) % p == 0)) ++ct;
  }
  cout << ct << endl;
}

```