#include <bits/stdc++.h>
using namespace std;
int prime;
int mod(int a, int n, int p) {
  if (n == 0) {
    return 1;
  } else if (n == 1) {
    return (a % p);
  }
  if (n % 2 == 0) {
    int res = mod(a, n / 2, p) * mod(a, n / 2, p);
    res %= p;
    return res;
  } else
    return (mod(a, n - 1, p) * (a % p)) % p;
}
int ispri(double num) {
  int tmp = (mod(num, prime - 1, prime) - 1) % prime;
  if (tmp % prime) {
    return 0;
  }
  for (int i = 1; i <= prime - 2; i++) {
    if (i == prime - 1) {
      continue;
    }
    int tmp = (mod(num, i, prime) - 1) % prime;
    if (tmp % prime == 0) {
      return 0;
    }
  }
  return 1;
}
int main() {
  while (cin >> prime) {
    int ans = 0;
    for (int i = 1; i < prime; i++) {
      if (ispri(i)) {
        ans++;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please formulate 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;
int prime;
int mod(int a, int n, int p) {
  if (n == 0) {
    return 1;
  } else if (n == 1) {
    return (a % p);
  }
  if (n % 2 == 0) {
    int res = mod(a, n / 2, p) * mod(a, n / 2, p);
    res %= p;
    return res;
  } else
    return (mod(a, n - 1, p) * (a % p)) % p;
}
int ispri(double num) {
  int tmp = (mod(num, prime - 1, prime) - 1) % prime;
  if (tmp % prime) {
    return 0;
  }
  for (int i = 1; i <= prime - 2; i++) {
    if (i == prime - 1) {
      continue;
    }
    int tmp = (mod(num, i, prime) - 1) % prime;
    if (tmp % prime == 0) {
      return 0;
    }
  }
  return 1;
}
int main() {
  while (cin >> prime) {
    int ans = 0;
    for (int i = 1; i < prime; i++) {
      if (ispri(i)) {
        ans++;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```