#include <bits/stdc++.h>
using namespace std;
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  long long p, count, finalcount = 0;
  cin >> p;
  for (int i = 1; i < p; ++i) {
    count = 0;
    for (int j = 1; j < p - 1; ++j) {
      if ((binpow(i, j, p) - 1 + p) % p != 0) {
        count++;
      }
    }
    if ((binpow(i, p - 1, p) - 1 + p) % p == 0) {
      count++;
    }
    if (count == p - 1) {
      finalcount++;
    }
  }
  cout << finalcount << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  long long p, count, finalcount = 0;
  cin >> p;
  for (int i = 1; i < p; ++i) {
    count = 0;
    for (int j = 1; j < p - 1; ++j) {
      if ((binpow(i, j, p) - 1 + p) % p != 0) {
        count++;
      }
    }
    if ((binpow(i, p - 1, p) - 1 + p) % p == 0) {
      count++;
    }
    if (count == p - 1) {
      finalcount++;
    }
  }
  cout << finalcount << endl;
  return 0;
}

```