#include <bits/stdc++.h>
int ask;
int check_root(int num, int cifang) {
  int i, k = 1;
  for (i = 0; i < cifang; ++i) {
    k *= num;
    k = k % ask;
  }
  if (k > 0)
    return k - 1;
  else
    return num - 1;
}
int check_root2(int num, int cifang) {
  int i, k = 1;
  for (i = 0; i < cifang; ++i) {
    k *= num;
    k = k % ask;
    if (k == 1) return 0;
  }
  return 1;
}
int main() {
  int i, j;
  int count = 0, rec = 0;
  scanf("%d", &ask);
  for (i = 1; i < ask; ++i) {
    count = 0;
    if (check_root(i, ask - 1) == 0) {
      if (check_root2(i, ask - 2) == 1) rec += 1;
    }
  }
  printf("%d\n", rec);
  return 0;
}

### Prompt

Please create a solution in CPP 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>
int ask;
int check_root(int num, int cifang) {
  int i, k = 1;
  for (i = 0; i < cifang; ++i) {
    k *= num;
    k = k % ask;
  }
  if (k > 0)
    return k - 1;
  else
    return num - 1;
}
int check_root2(int num, int cifang) {
  int i, k = 1;
  for (i = 0; i < cifang; ++i) {
    k *= num;
    k = k % ask;
    if (k == 1) return 0;
  }
  return 1;
}
int main() {
  int i, j;
  int count = 0, rec = 0;
  scanf("%d", &ask);
  for (i = 1; i < ask; ++i) {
    count = 0;
    if (check_root(i, ask - 1) == 0) {
      if (check_root2(i, ask - 2) == 1) rec += 1;
    }
  }
  printf("%d\n", rec);
  return 0;
}

```