#include <bits/stdc++.h>
int main() {
  int p, count(0);
  scanf("%d", &p);
  for (int n = 1; n < p; ++n) {
    bool ok = true;
    int x = 1;
    for (int i = 1; i < p - 1; ++i) {
      x *= n;
      x %= p;
      if (x == 1) {
        ok = false;
        break;
      }
    }
    if (ok) {
      x *= n;
      x %= p;
      if (x == 1) {
        count += 1;
      }
    }
  }
  printf("%d\n", count);
  return 0;
}

### Prompt

Your task is to create 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>
int main() {
  int p, count(0);
  scanf("%d", &p);
  for (int n = 1; n < p; ++n) {
    bool ok = true;
    int x = 1;
    for (int i = 1; i < p - 1; ++i) {
      x *= n;
      x %= p;
      if (x == 1) {
        ok = false;
        break;
      }
    }
    if (ok) {
      x *= n;
      x %= p;
      if (x == 1) {
        count += 1;
      }
    }
  }
  printf("%d\n", count);
  return 0;
}

```