#include <bits/stdc++.h>
const long long MOD = 1e9 + 7;
using namespace std;
void solve() {
  long long p;
  cin >> p;
  long long c = 0;
  for (long long i = 1; i <= p; i++) {
    bool cond = true;
    long long x = i % p;
    long long t = 1;
    for (long long j = 1; j < p - 1; j++) {
      t = (t * x) % p;
      if (((t - 1) % p) == 0) {
        cond = false;
        break;
      }
    }
    if (cond) {
      long long z = (t * x) % p;
      if ((z - 1) % p == 0) {
        c++;
      }
    }
  }
  cout << c;
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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>
const long long MOD = 1e9 + 7;
using namespace std;
void solve() {
  long long p;
  cin >> p;
  long long c = 0;
  for (long long i = 1; i <= p; i++) {
    bool cond = true;
    long long x = i % p;
    long long t = 1;
    for (long long j = 1; j < p - 1; j++) {
      t = (t * x) % p;
      if (((t - 1) % p) == 0) {
        cond = false;
        break;
      }
    }
    if (cond) {
      long long z = (t * x) % p;
      if ((z - 1) % p == 0) {
        c++;
      }
    }
  }
  cout << c;
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```