#include <bits/stdc++.h>
using namespace std;
void solve() {
  int p;
  cin >> p;
  int count = 0, k = 0;
  for (int x = 1; x < p; x++) {
    k = 1;
    for (int i = 1; i < p - 1; i++) {
      k = (x * k) % p;
      if (k == 1) {
        goto abc;
      }
    }
    if (k * x % p == 1) {
      count++;
    }
  abc:;
  }
  cout << count;
  return;
}
int main() { solve(); }

### Prompt

In cpp, your task is to solve 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;
void solve() {
  int p;
  cin >> p;
  int count = 0, k = 0;
  for (int x = 1; x < p; x++) {
    k = 1;
    for (int i = 1; i < p - 1; i++) {
      k = (x * k) % p;
      if (k == 1) {
        goto abc;
      }
    }
    if (k * x % p == 1) {
      count++;
    }
  abc:;
  }
  cout << count;
  return;
}
int main() { solve(); }

```