#include <bits/stdc++.h>
using namespace std;
int mod;
long long qp(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
int main() {
  int p;
  cin >> p;
  mod = p;
  int ans = 0;
  for (int x = 1; x < p; ++x) {
    long long cur = 1;
    int j;
    for (j = 1; j <= p - 2; ++j) {
      cur = cur * x % mod;
      if ((cur + mod - 1) % p == 0) {
        break;
      }
    }
    cur = cur * x % mod;
    if (j == p - 1 && (cur + mod - 1) % p == 0) ans++;
  }
  cout << ans << endl;
}

### Prompt

Please provide a Cpp coded solution 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;
int mod;
long long qp(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
int main() {
  int p;
  cin >> p;
  mod = p;
  int ans = 0;
  for (int x = 1; x < p; ++x) {
    long long cur = 1;
    int j;
    for (j = 1; j <= p - 2; ++j) {
      cur = cur * x % mod;
      if ((cur + mod - 1) % p == 0) {
        break;
      }
    }
    cur = cur * x % mod;
    if (j == p - 1 && (cur + mod - 1) % p == 0) ans++;
  }
  cout << ans << endl;
}

```