#include <bits/stdc++.h>
using namespace std;
int mod(int a, int b, int c) {
  if (b == 0) return 1;
  if (b == 1) return a % c;
  if (b & 1)
    return (a * mod(a, b - 1, c)) % c;
  else {
    int tmp = mod(a, b >> 1, c);
    return (tmp * tmp) % c;
  }
}
bool chk(int x, int p) {
  if (mod(x, p - 1, p) != 1) return false;
  for (int i = 1; i <= p - 2; i++) {
    int tmp = mod(x, i, p);
    if (tmp == 1) return false;
  }
  return true;
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    int ans = 0;
    for (int i = 1; i < n; i++)
      if (chk(i, n)) ans++;
    printf("%d\n", ans);
  }
  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>
using namespace std;
int mod(int a, int b, int c) {
  if (b == 0) return 1;
  if (b == 1) return a % c;
  if (b & 1)
    return (a * mod(a, b - 1, c)) % c;
  else {
    int tmp = mod(a, b >> 1, c);
    return (tmp * tmp) % c;
  }
}
bool chk(int x, int p) {
  if (mod(x, p - 1, p) != 1) return false;
  for (int i = 1; i <= p - 2; i++) {
    int tmp = mod(x, i, p);
    if (tmp == 1) return false;
  }
  return true;
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    int ans = 0;
    for (int i = 1; i < n; i++)
      if (chk(i, n)) ans++;
    printf("%d\n", ans);
  }
  return 0;
}

```