#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, j, k = 0, p, a[2001], len = 0;
  cin >> p;
  if (p == 2) {
    cout << "1" << endl;
    return 0;
  }
  for (i = 2; i < p; i++) {
    j = i;
    for (k = 1; k < p - 1; k++) {
      if ((j - 1) % p == 0) k = p;
      j = (j * i) % p;
    }
    if (k == p - 1 && (j - 1) % p == 0) len++;
  }
  cout << len << endl;
  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 i, j, k = 0, p, a[2001], len = 0;
  cin >> p;
  if (p == 2) {
    cout << "1" << endl;
    return 0;
  }
  for (i = 2; i < p; i++) {
    j = i;
    for (k = 1; k < p - 1; k++) {
      if ((j - 1) % p == 0) k = p;
      j = (j * i) % p;
    }
    if (k == p - 1 && (j - 1) % p == 0) len++;
  }
  cout << len << endl;
  return 0;
}

```