#include <bits/stdc++.h>
using namespace std;
long long modPow(long long b, long long e, long long mod) {
  long long result = 1;
  b %= mod;
  while (e) {
    if (e & 1) result = ((long long)result * b) % mod;
    b = (b * b) % mod;
    e >>= 1;
  }
  return result;
}
int main(int argc, char const* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int p;
  cin >> p;
  long long sol = 0;
  for (int i = 1; i <= p; ++i)
    for (int j = 1; j <= p - 1; ++j)
      if ((modPow(i, j, p) - 1) % p == 0 && j != p - 1)
        break;
      else if (j == p - 1 && (modPow(i, j, p) - 1) % p == 0)
        ++sol;
  cout << sol << '\n';
  return 0;
}

### Prompt

Generate 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;
long long modPow(long long b, long long e, long long mod) {
  long long result = 1;
  b %= mod;
  while (e) {
    if (e & 1) result = ((long long)result * b) % mod;
    b = (b * b) % mod;
    e >>= 1;
  }
  return result;
}
int main(int argc, char const* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int p;
  cin >> p;
  long long sol = 0;
  for (int i = 1; i <= p; ++i)
    for (int j = 1; j <= p - 1; ++j)
      if ((modPow(i, j, p) - 1) % p == 0 && j != p - 1)
        break;
      else if (j == p - 1 && (modPow(i, j, p) - 1) % p == 0)
        ++sol;
  cout << sol << '\n';
  return 0;
}

```