#include <bits/stdc++.h>
using namespace std;
long long fast_pow(long long x, long long p, long long P) {
  if (p == 0) return 1;
  long long res = fast_pow(x, p / 2, P);
  if (p & 1)
    return (res * res * x) % P;
  else
    return (res * res) % P;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long p, ans = 0;
  cin >> p;
  for (long long x = 1; x < p; x++) {
    bool b = true;
    for (long long i = 1; i <= p - 2; i++)
      if ((fast_pow(x, i, p) - 1) % p == 0) b = false;
    if (b && (fast_pow(x, p - 1, p) - 1) % p == 0) 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 fast_pow(long long x, long long p, long long P) {
  if (p == 0) return 1;
  long long res = fast_pow(x, p / 2, P);
  if (p & 1)
    return (res * res * x) % P;
  else
    return (res * res) % P;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long p, ans = 0;
  cin >> p;
  for (long long x = 1; x < p; x++) {
    bool b = true;
    for (long long i = 1; i <= p - 2; i++)
      if ((fast_pow(x, i, p) - 1) % p == 0) b = false;
    if (b && (fast_pow(x, p - 1, p) - 1) % p == 0) ans++;
  }
  cout << ans;
  return 0;
}

```