#include <bits/stdc++.h>
using namespace std;
vector<int> perm(5, 1);
long long fastpow(long long base, long long y, long long p) {
  if (y == 1) return base;
  long long x = fastpow(base, y / 2, p);
  if (!(y % 2)) {
    return (x * x) % p;
  } else {
    return (x * x * base) % p;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int p, c = 0;
  cin >> p;
  for (int i = 1; i < p; i++) {
    bool k = 1;
    for (int j = 1; j <= p - 2; j++) {
      if ((fastpow(i, j, p) - 1) % p == 0) {
        k = 0;
        break;
      }
    }
    if ((fastpow(i, p - 1, p) - 1) % p == 0 && k) {
      c++;
    }
  }
  cout << c;
  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;
vector<int> perm(5, 1);
long long fastpow(long long base, long long y, long long p) {
  if (y == 1) return base;
  long long x = fastpow(base, y / 2, p);
  if (!(y % 2)) {
    return (x * x) % p;
  } else {
    return (x * x * base) % p;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int p, c = 0;
  cin >> p;
  for (int i = 1; i < p; i++) {
    bool k = 1;
    for (int j = 1; j <= p - 2; j++) {
      if ((fastpow(i, j, p) - 1) % p == 0) {
        k = 0;
        break;
      }
    }
    if ((fastpow(i, p - 1, p) - 1) % p == 0 && k) {
      c++;
    }
  }
  cout << c;
  return 0;
}

```