#include <bits/stdc++.h>
using namespace std;
int p, nr, i, j;
int put(int x, int y) {
  int rez = 1;
  while (y > 0)
    if (y % 2 == 0) {
      x *= x;
      x %= p;
      y /= 2;
    } else {
      rez *= x;
      rez %= p;
      --y;
    }
  return (rez);
}
int main(void) {
  cin >> p;
  for (i = 1; i < p; ++i) {
    bool ok1 = 1;
    for (j = 1; j < p - 1; ++j)
      if (put(i, j) == 1) {
        ok1 = 0;
        break;
      }
    if (ok1 == 1 && put(i, p - 1) == 1) ++nr;
  }
  cout << nr;
  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 p, nr, i, j;
int put(int x, int y) {
  int rez = 1;
  while (y > 0)
    if (y % 2 == 0) {
      x *= x;
      x %= p;
      y /= 2;
    } else {
      rez *= x;
      rez %= p;
      --y;
    }
  return (rez);
}
int main(void) {
  cin >> p;
  for (i = 1; i < p; ++i) {
    bool ok1 = 1;
    for (j = 1; j < p - 1; ++j)
      if (put(i, j) == 1) {
        ok1 = 0;
        break;
      }
    if (ok1 == 1 && put(i, p - 1) == 1) ++nr;
  }
  cout << nr;
  return (0);
}

```