#include <bits/stdc++.h>
using namespace std;
int p;
int AnswerProblem;
bool valid(int num) {
  bool is_bad = false;
  long long nowst = 1;
  for (int pp = 1; pp <= (p - 2); pp++) {
    nowst *= num;
    nowst %= p;
    int nowst1 = nowst - 1;
    if (nowst1 < 0) nowst1 += p;
    if (nowst1 == 0) is_bad = true;
  }
  nowst *= p;
  if (is_bad) return false;
  if (nowst % p) return false;
  return true;
}
int main() {
  cin >> p;
  for (int x = 1; x < p; x++) {
    if (valid(x)) {
      AnswerProblem++;
    }
  }
  cout << AnswerProblem << endl;
  return 0;
}

### 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 p;
int AnswerProblem;
bool valid(int num) {
  bool is_bad = false;
  long long nowst = 1;
  for (int pp = 1; pp <= (p - 2); pp++) {
    nowst *= num;
    nowst %= p;
    int nowst1 = nowst - 1;
    if (nowst1 < 0) nowst1 += p;
    if (nowst1 == 0) is_bad = true;
  }
  nowst *= p;
  if (is_bad) return false;
  if (nowst % p) return false;
  return true;
}
int main() {
  cin >> p;
  for (int x = 1; x < p; x++) {
    if (valid(x)) {
      AnswerProblem++;
    }
  }
  cout << AnswerProblem << endl;
  return 0;
}

```