#include <bits/stdc++.h>
int cal(int max) {
  int i, sum = 0, j, k, da;
  for (k = 1; k < max; k++) {
    da = k;
    int flag = 1;
    for (j = 1; j < max - 1; j++) {
      if ((da - 1) % max == 0) {
        flag = 0;
        break;
      } else
        da = da * k % max;
    }
    if ((flag == 1) && ((da - 1) % max == 0)) sum++;
  }
  return sum;
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    printf("%d\n", cal(n));
  }
  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>
int cal(int max) {
  int i, sum = 0, j, k, da;
  for (k = 1; k < max; k++) {
    da = k;
    int flag = 1;
    for (j = 1; j < max - 1; j++) {
      if ((da - 1) % max == 0) {
        flag = 0;
        break;
      } else
        da = da * k % max;
    }
    if ((flag == 1) && ((da - 1) % max == 0)) sum++;
  }
  return sum;
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    printf("%d\n", cal(n));
  }
  return 0;
}

```