#include <bits/stdc++.h>
long long fastpower(long long base, long long power, long long mod) {
  if (!power) return 1;
  long long left = fastpower(base, power / 2, mod);
  left *= left;
  left %= mod;
  if (power % 2 != 0)
    return (left * (base % mod)) % mod;
  else if (power % 2 == 0)
    return left;
}
using namespace std;
int main(void) {
  int p, cntr = 0;
  cin >> p;
  for (long long x = 1; x < p; x++) {
    bool div = true;
    for (int pow = 1; pow < (p - 1); ++pow) {
      if ((fastpower(x, pow, p) - 1) % p == 0) {
        div = false;
        break;
      }
    }
    if (div && (fastpower(x, p - 1, p) - 1) % p == 0) cntr++;
  }
  cout << cntr << endl;
}

### 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>
long long fastpower(long long base, long long power, long long mod) {
  if (!power) return 1;
  long long left = fastpower(base, power / 2, mod);
  left *= left;
  left %= mod;
  if (power % 2 != 0)
    return (left * (base % mod)) % mod;
  else if (power % 2 == 0)
    return left;
}
using namespace std;
int main(void) {
  int p, cntr = 0;
  cin >> p;
  for (long long x = 1; x < p; x++) {
    bool div = true;
    for (int pow = 1; pow < (p - 1); ++pow) {
      if ((fastpower(x, pow, p) - 1) % p == 0) {
        div = false;
        break;
      }
    }
    if (div && (fastpower(x, p - 1, p) - 1) % p == 0) cntr++;
  }
  cout << cntr << endl;
}

```