#include <bits/stdc++.h>
using namespace std;
int INP, AM, REACHEOF;
char BUF[(1 << 12) + 1], *inp = BUF;
const long double PI = acos((long double)-1.0);
bool check(int x, int p) {
  int cur = 1;
  for (int i = (1), _b = (p - 2); i <= _b; i++) {
    cur = (cur * x) % p;
    if (cur == 1) return false;
  }
  cur = (cur * x) % p;
  return cur == 1;
}
int main() {
  int p;
  while (cin >> p) {
    int res = 0;
    for (int x = 0, _a = (p); x < _a; x++)
      if (check(x, p)) ++res;
    cout << res << endl;
  }
  return 0;
}

### Prompt

Please formulate 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 INP, AM, REACHEOF;
char BUF[(1 << 12) + 1], *inp = BUF;
const long double PI = acos((long double)-1.0);
bool check(int x, int p) {
  int cur = 1;
  for (int i = (1), _b = (p - 2); i <= _b; i++) {
    cur = (cur * x) % p;
    if (cur == 1) return false;
  }
  cur = (cur * x) % p;
  return cur == 1;
}
int main() {
  int p;
  while (cin >> p) {
    int res = 0;
    for (int x = 0, _a = (p); x < _a; x++)
      if (check(x, p)) ++res;
    cout << res << endl;
  }
  return 0;
}

```