#include <bits/stdc++.h>
long long int phi[2002];
int main() {
  int i, j, N = 2000, n;
  for (i = 1; i <= N; i++) phi[i] = i;
  for (i = 2; i <= N; i++) {
    if (phi[i] == i)
      for (j = i; j <= N; j = j + i) phi[j] = (phi[j] / i) * (i - 1);
  }
  scanf("%d", &n);
  printf("%I64d\n", phi[phi[n]]);
}

### Prompt

Create a solution in CPP for 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>
long long int phi[2002];
int main() {
  int i, j, N = 2000, n;
  for (i = 1; i <= N; i++) phi[i] = i;
  for (i = 2; i <= N; i++) {
    if (phi[i] == i)
      for (j = i; j <= N; j = j + i) phi[j] = (phi[j] / i) * (i - 1);
  }
  scanf("%d", &n);
  printf("%I64d\n", phi[phi[n]]);
}

```