#include <bits/stdc++.h>
using ll = long long;
using namespace std;
int p;
ll power(ll base, ll exp) {
  if (exp == 0) return 1;
  if (exp == 1) return base % p;
  ll t = power(base, exp / 2);
  t = (t * t) % p;
  if (exp % 2 == 0)
    return t;
  else
    return ((base % p) * t) % p;
}
int main() {
  cin >> p;
  int ans = 0;
  for (int i = 1; i < p; i++) {
    bool t = true;
    for (int j = 1; j <= p - 2; j++) {
      if (power(i, j) == 1) {
        t = false;
        break;
      }
    }
    if (t && power(i, p - 1) == 1) ans++;
  }
  cout << ans;
}

### 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 ll = long long;
using namespace std;
int p;
ll power(ll base, ll exp) {
  if (exp == 0) return 1;
  if (exp == 1) return base % p;
  ll t = power(base, exp / 2);
  t = (t * t) % p;
  if (exp % 2 == 0)
    return t;
  else
    return ((base % p) * t) % p;
}
int main() {
  cin >> p;
  int ans = 0;
  for (int i = 1; i < p; i++) {
    bool t = true;
    for (int j = 1; j <= p - 2; j++) {
      if (power(i, j) == 1) {
        t = false;
        break;
      }
    }
    if (t && power(i, p - 1) == 1) ans++;
  }
  cout << ans;
}

```