#include <bits/stdc++.h>
using namespace std;
int p;
int power(int x, int y) {
  if (y == 0) {
    return 1;
  }
  int temp = power(x, y >> 1);
  if (y & 1) {
    return ((((long long)temp * temp) % p) * x) % p;
  }
  return ((long long)temp * temp) % p;
}
bool cntIsOne(int x) {
  for (int i = 1; i < p - 1; i++) {
    if (power(x, i) == 1) {
      return false;
    }
  }
  return power(x, p - 1) == 1;
}
int main() {
  scanf("%d", &p);
  int res = 0;
  for (int i = 1; i < p; i++) {
    res += cntIsOne(i);
  }
  printf("%d", res);
  return 0;
}

### Prompt

Create a solution in CPP for 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 p;
int power(int x, int y) {
  if (y == 0) {
    return 1;
  }
  int temp = power(x, y >> 1);
  if (y & 1) {
    return ((((long long)temp * temp) % p) * x) % p;
  }
  return ((long long)temp * temp) % p;
}
bool cntIsOne(int x) {
  for (int i = 1; i < p - 1; i++) {
    if (power(x, i) == 1) {
      return false;
    }
  }
  return power(x, p - 1) == 1;
}
int main() {
  scanf("%d", &p);
  int res = 0;
  for (int i = 1; i < p; i++) {
    res += cntIsOne(i);
  }
  printf("%d", res);
  return 0;
}

```