#include <bits/stdc++.h>
int intcmp(const void *f, const void *s) { return (*(int *)f - *(int *)s); }
int gcd(int a, int b) { return ((b == 0) ? a : gcd(b, a % b)); }
int preprocess() { return 0; }
int main() {
  int i, j, n;
  int p;
  int ans = 0;
  scanf("%d", &p);
  for (i = 1; i < p; ++i) {
    int res = 1;
    for (j = 1; j < p; ++j) {
      res = (res * i) % p;
      if (res == 1) break;
    }
    if (j == p - 1) ans++;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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>
int intcmp(const void *f, const void *s) { return (*(int *)f - *(int *)s); }
int gcd(int a, int b) { return ((b == 0) ? a : gcd(b, a % b)); }
int preprocess() { return 0; }
int main() {
  int i, j, n;
  int p;
  int ans = 0;
  scanf("%d", &p);
  for (i = 1; i < p; ++i) {
    int res = 1;
    for (j = 1; j < p; ++j) {
      res = (res * i) % p;
      if (res == 1) break;
    }
    if (j == p - 1) ans++;
  }
  printf("%d\n", ans);
  return 0;
}

```