#include <bits/stdc++.h>
using namespace std;
string inttostring(int n) {
  stringstream a;
  a << n;
  return a.str();
}
int stringtoint(string A) {
  istringstream a(A);
  int p;
  a >> p;
  return p;
}
int p;
long long pow(int n, int k) {
  if (k == 1)
    return n;
  else {
    long long c = pow(n, k / 2);
    c = (c * c) % p;
    if (k % 2 == 0) {
      return c;
    } else {
      return (c * n) % p;
    }
  }
}
int main() {
  cin >> p;
  int res = 0;
  bool t = false;
  for (int n = 1; n < p; n++) {
    t = false;
    for (int i = 1; i < p - 1; i++) {
      if ((pow(n, i) - 1) % p == 0) {
        t = true;
        break;
      }
    }
    if (!t) {
      if ((pow(n, p - 1) - 1) % p == 0) res++;
    }
  }
  cout << res << "\n";
}

### 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;
string inttostring(int n) {
  stringstream a;
  a << n;
  return a.str();
}
int stringtoint(string A) {
  istringstream a(A);
  int p;
  a >> p;
  return p;
}
int p;
long long pow(int n, int k) {
  if (k == 1)
    return n;
  else {
    long long c = pow(n, k / 2);
    c = (c * c) % p;
    if (k % 2 == 0) {
      return c;
    } else {
      return (c * n) % p;
    }
  }
}
int main() {
  cin >> p;
  int res = 0;
  bool t = false;
  for (int n = 1; n < p; n++) {
    t = false;
    for (int i = 1; i < p - 1; i++) {
      if ((pow(n, i) - 1) % p == 0) {
        t = true;
        break;
      }
    }
    if (!t) {
      if ((pow(n, p - 1) - 1) % p == 0) res++;
    }
  }
  cout << res << "\n";
}

```