#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
int p;
int pow_mod(int base, int power) {
  if (power <= 0) return 1;
  if (power == 1) return base;
  if (power & 1) return (base * pow_mod(base, power - 1) % p) % p;
  return pow_mod((base * base) % p, power >> 1) % p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> p;
  int res = 0;
  for (__typeof(p) i = (1) - ((1) > (p)); i != (p) - ((1) > (p));
       i += 1 - 2 * ((1) > (p))) {
    bool can = true;
    for (__typeof(p - 1) j = (1) - ((1) > (p - 1));
         j != (p - 1) - ((1) > (p - 1)); j += 1 - 2 * ((1) > (p - 1))) {
      if (pow_mod(i, j) == 1) {
        can = false;
        break;
      }
    }
    can &= pow_mod(i, p - 1) == 1;
    res += can;
  }
  cout << res;
  return 0;
}

### Prompt

Generate 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;
const long long mod = 1000000007;
int p;
int pow_mod(int base, int power) {
  if (power <= 0) return 1;
  if (power == 1) return base;
  if (power & 1) return (base * pow_mod(base, power - 1) % p) % p;
  return pow_mod((base * base) % p, power >> 1) % p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> p;
  int res = 0;
  for (__typeof(p) i = (1) - ((1) > (p)); i != (p) - ((1) > (p));
       i += 1 - 2 * ((1) > (p))) {
    bool can = true;
    for (__typeof(p - 1) j = (1) - ((1) > (p - 1));
         j != (p - 1) - ((1) > (p - 1)); j += 1 - 2 * ((1) > (p - 1))) {
      if (pow_mod(i, j) == 1) {
        can = false;
        break;
      }
    }
    can &= pow_mod(i, p - 1) == 1;
    res += can;
  }
  cout << res;
  return 0;
}

```