#include <bits/stdc++.h>
using namespace std;
bool ok(int x, int p) {
  bool good = true;
  int cur = 1;
  for (int i = 1; i < (p - 1); i++) {
    cur *= x;
    cur %= p;
    int curnum = ((cur - 1) + p) % p;
    if (curnum == 0) return false;
  }
  cur *= x;
  cur %= p;
  cur = ((cur - 1) + p) % p;
  if (cur == 0) return true;
  return false;
}
int solve(int p) {
  int ret = 0;
  for (int i = 1; i < p; i++) {
    if (ok(i, p)) ret++;
  }
  return ret;
}
int main() {
  int p = 0;
  cin >> p;
  int res = solve(p);
  cout << res << endl;
  return 0;
}

### 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;
bool ok(int x, int p) {
  bool good = true;
  int cur = 1;
  for (int i = 1; i < (p - 1); i++) {
    cur *= x;
    cur %= p;
    int curnum = ((cur - 1) + p) % p;
    if (curnum == 0) return false;
  }
  cur *= x;
  cur %= p;
  cur = ((cur - 1) + p) % p;
  if (cur == 0) return true;
  return false;
}
int solve(int p) {
  int ret = 0;
  for (int i = 1; i < p; i++) {
    if (ok(i, p)) ret++;
  }
  return ret;
}
int main() {
  int p = 0;
  cin >> p;
  int res = solve(p);
  cout << res << endl;
  return 0;
}

```