#include <bits/stdc++.h>
using namespace std;
int p;
int x;
void input();
void solve();
int main() {
  input();
  solve();
  return 0;
}
void input() { scanf("%d", &p); }
void solve() {
  int i;
  int j;
  int ans = 0;
  int fl = 0;
  int num = 1;
  int g;
  for (i = 1; i < p; i++) {
    fl = 0;
    num = 1;
    g = i;
    for (j = 1; j <= p - 1; j++) {
      num = num * g;
      num = num % p;
      if (num == 1 && j != p - 1) {
        fl = 1;
        break;
      }
      if (num == 1) break;
    }
    if (fl == 0 && num == 1) ans++;
  }
  printf("%d\n", ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 x;
void input();
void solve();
int main() {
  input();
  solve();
  return 0;
}
void input() { scanf("%d", &p); }
void solve() {
  int i;
  int j;
  int ans = 0;
  int fl = 0;
  int num = 1;
  int g;
  for (i = 1; i < p; i++) {
    fl = 0;
    num = 1;
    g = i;
    for (j = 1; j <= p - 1; j++) {
      num = num * g;
      num = num % p;
      if (num == 1 && j != p - 1) {
        fl = 1;
        break;
      }
      if (num == 1) break;
    }
    if (fl == 0 && num == 1) ans++;
  }
  printf("%d\n", ans);
}

```