#include <bits/stdc++.h>
using namespace std;
long long powmod(int n, int b, int p) {
  int res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = (res * (long long)n) % p;
    n = ((n % p) * (n % p)) % p;
  }
  return res;
}
int main() {
  int p, count = 0;
  cin >> p;
  if (p == 2 || p == 3) {
    cout << 1 << "\n";
    return 0;
  }
  for (int i = 2; i < p - 1; i++) {
    int chk = 1;
    for (int j = 2; j < p - 1; j++)
      if ((powmod(i, j, p) - 1) % p == 0) {
        chk = 0;
        break;
      }
    if (powmod(i, p - 1, p) and chk == 1) count++;
  }
  cout << count << "\n";
  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;
long long powmod(int n, int b, int p) {
  int res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = (res * (long long)n) % p;
    n = ((n % p) * (n % p)) % p;
  }
  return res;
}
int main() {
  int p, count = 0;
  cin >> p;
  if (p == 2 || p == 3) {
    cout << 1 << "\n";
    return 0;
  }
  for (int i = 2; i < p - 1; i++) {
    int chk = 1;
    for (int j = 2; j < p - 1; j++)
      if ((powmod(i, j, p) - 1) % p == 0) {
        chk = 0;
        break;
      }
    if (powmod(i, p - 1, p) and chk == 1) count++;
  }
  cout << count << "\n";
  return 0;
}

```