#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    long long n, cnt = 0, flag = 0;
    cin >> n;
    for (long long i = 1; i < n; i++) {
      flag = 0;
      for (long long j = 1; j <= n - 2; j++) {
        if ((power(i, j, n) - 1) % n == 0) {
          flag = 1;
          break;
        }
      }
      if ((power(i, n - 1, n) - 1) % n != 0) {
        flag = 1;
      }
      if (flag == 0) cnt++;
    }
    cout << cnt;
  }
}

### Prompt

Please create a solution in CPP 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;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    long long n, cnt = 0, flag = 0;
    cin >> n;
    for (long long i = 1; i < n; i++) {
      flag = 0;
      for (long long j = 1; j <= n - 2; j++) {
        if ((power(i, j, n) - 1) % n == 0) {
          flag = 1;
          break;
        }
      }
      if ((power(i, n - 1, n) - 1) % n != 0) {
        flag = 1;
      }
      if (flag == 0) cnt++;
    }
    cout << cnt;
  }
}

```