#include <bits/stdc++.h>
using namespace std;
const double pi = 2 * acos(0.0);
int OO = 0x3f3f3f3f;
long long mod;
long long fp(long long b, long long p) {
  if (p == 0) return 1;
  long long ans = fp(b, p / 2);
  ans = (ans * ans) % mod;
  if (p % 2) ans = (ans * (b % mod)) % mod;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long p, ans, cnt = 0;
  cin >> p;
  mod = p;
  for (int i = 1; i < p; i++) {
    bool Flag = 1;
    for (int j = 1; j < p - 1; j++) {
      ans = fp(i, j) - 1;
      if (!ans) {
        Flag = 0;
        break;
      }
    }
    if (Flag && fp(i, p - 1) - 1 == 0) ++cnt;
  }
  cout << cnt;
  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;
const double pi = 2 * acos(0.0);
int OO = 0x3f3f3f3f;
long long mod;
long long fp(long long b, long long p) {
  if (p == 0) return 1;
  long long ans = fp(b, p / 2);
  ans = (ans * ans) % mod;
  if (p % 2) ans = (ans * (b % mod)) % mod;
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long p, ans, cnt = 0;
  cin >> p;
  mod = p;
  for (int i = 1; i < p; i++) {
    bool Flag = 1;
    for (int j = 1; j < p - 1; j++) {
      ans = fp(i, j) - 1;
      if (!ans) {
        Flag = 0;
        break;
      }
    }
    if (Flag && fp(i, p - 1) - 1 == 0) ++cnt;
  }
  cout << cnt;
  return 0;
}

```