#include <bits/stdc++.h>
using namespace std;
long long binpowMOD(long long a, long long n, int MOD) {
  long long res = 1;
  a %= MOD;
  while (n) {
    if (n & 1) {
      res *= a;
      res %= MOD;
    }
    a *= a;
    a %= MOD;
    n >>= 1;
  }
  return res;
}
bool isgood(long long x, int p) {
  for (int i = 1; i <= p - 2; i++) {
    if ((binpowMOD(x, i, p) + p - 1) % p == 0) return false;
  }
  if ((binpowMOD(x, p - 1, p) + p - 1) % p == 0) return true;
  return false;
}
int main() {
  int p, ans = 0;
  cin >> p;
  for (int i = 1; i < p; i++) {
    if (isgood(i, p)) ans++;
  }
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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 binpowMOD(long long a, long long n, int MOD) {
  long long res = 1;
  a %= MOD;
  while (n) {
    if (n & 1) {
      res *= a;
      res %= MOD;
    }
    a *= a;
    a %= MOD;
    n >>= 1;
  }
  return res;
}
bool isgood(long long x, int p) {
  for (int i = 1; i <= p - 2; i++) {
    if ((binpowMOD(x, i, p) + p - 1) % p == 0) return false;
  }
  if ((binpowMOD(x, p - 1, p) + p - 1) % p == 0) return true;
  return false;
}
int main() {
  int p, ans = 0;
  cin >> p;
  for (int i = 1; i < p; i++) {
    if (isgood(i, p)) ans++;
  }
  cout << ans;
  return 0;
}

```