#include <bits/stdc++.h>
using namespace std;
int check(int x, int p) {
  int j = x;
  for (int i = 0; i < p - 2; ++i, j = (j * x) % p) {
    if (j == 1) return 0;
  }
  return j == 1;
}
int main() {
  int p;
  scanf("%d", &p);
  int count = 0;
  if (p == 2)
    printf("1");
  else {
    for (int i = 2; i < p; ++i) {
      count += check(i, p);
    }
    printf("%d", count);
  }
  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 check(int x, int p) {
  int j = x;
  for (int i = 0; i < p - 2; ++i, j = (j * x) % p) {
    if (j == 1) return 0;
  }
  return j == 1;
}
int main() {
  int p;
  scanf("%d", &p);
  int count = 0;
  if (p == 2)
    printf("1");
  else {
    for (int i = 2; i < p; ++i) {
      count += check(i, p);
    }
    printf("%d", count);
  }
  return 0;
}

```