#include <bits/stdc++.h>
using namespace std;
bool is_primitive_root(int x, int p) {
  int z = 1;
  for (int y = 0; y < p - 2; ++y) {
    z = (z * x) % p;
    if (z == 1) return false;
  }
  z = (z * x) % p;
  return z == 1;
}
void solve(istream &in, ostream &out) {
  int p;
  in >> p;
  int c = 0;
  for (int x = 1; x < p; ++x) {
    if (is_primitive_root(x, p)) {
      ++c;
    }
  }
  out << c << endl;
}
int main() {
  solve(cin, cout);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
bool is_primitive_root(int x, int p) {
  int z = 1;
  for (int y = 0; y < p - 2; ++y) {
    z = (z * x) % p;
    if (z == 1) return false;
  }
  z = (z * x) % p;
  return z == 1;
}
void solve(istream &in, ostream &out) {
  int p;
  in >> p;
  int c = 0;
  for (int x = 1; x < p; ++x) {
    if (is_primitive_root(x, p)) {
      ++c;
    }
  }
  out << c << endl;
}
int main() {
  solve(cin, cout);
  return 0;
}

```