#include <bits/stdc++.h>
using namespace std;
long p, cur, i, k, sum = 0, ind;
int main() {
  scanf("%li", &p);
  for (i = 1; i < p; i++) {
    cur = 1;
    ind = 0;
    for (k = 1; k < (p - 1); k++) {
      cur *= i;
      cur %= p;
      if (cur == 1) {
        ind = 1;
        break;
      }
    }
    cur *= i;
    cur %= p;
    if (cur != 1) {
      ind = 1;
    }
    if (ind == 0) {
      ++sum;
    }
  }
  printf("%li", sum);
  return 0;
}

### 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 p, cur, i, k, sum = 0, ind;
int main() {
  scanf("%li", &p);
  for (i = 1; i < p; i++) {
    cur = 1;
    ind = 0;
    for (k = 1; k < (p - 1); k++) {
      cur *= i;
      cur %= p;
      if (cur == 1) {
        ind = 1;
        break;
      }
    }
    cur *= i;
    cur %= p;
    if (cur != 1) {
      ind = 1;
    }
    if (ind == 0) {
      ++sum;
    }
  }
  printf("%li", sum);
  return 0;
}

```