#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, p, c = 1, sum = 0, flag;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    p = i;
    flag = 0;
    c = 1;
    while (true) {
      if (p % n == 1 && c < (n - 1)) {
        flag = 1;
        break;
      }
      if (p % n == 1 && c == (n - 1)) {
        flag = 2;
        break;
      }
      if (c > (n - 1)) break;
      p = (i * p) % n;
      c++;
    }
    if (flag == 2) sum++;
  }
  printf("%d", sum);
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
int main() {
  int n, p, c = 1, sum = 0, flag;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    p = i;
    flag = 0;
    c = 1;
    while (true) {
      if (p % n == 1 && c < (n - 1)) {
        flag = 1;
        break;
      }
      if (p % n == 1 && c == (n - 1)) {
        flag = 2;
        break;
      }
      if (c > (n - 1)) break;
      p = (i * p) % n;
      c++;
    }
    if (flag == 2) sum++;
  }
  printf("%d", sum);
  return 0;
}

```