#include <bits/stdc++.h>
int ModExp(int n, int m, int p) {
  int k = 1;
  n %= p;
  while (m != 1) {
    if (m & 1) {
      k = (k * n) % p;
    }
    n = (n * n) % p;
    m >>= 1;
  }
  return (n * k) % p;
}
int Judge(int x, int p) {
  int i;
  if (ModExp(x, p - 1, p) != 1) {
    return 0;
  }
  for (i = 1; i <= p - 2; ++i) {
    if (ModExp(x, i, p) == 1) {
      return 0;
    }
  }
  return 1;
}
int main() {
  int P, x, Sum;
  scanf("%d", &P);
  for (Sum = 0, x = 1; x < P; ++x) {
    Sum += Judge(x, P);
  }
  printf("%d\n", Sum);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 ModExp(int n, int m, int p) {
  int k = 1;
  n %= p;
  while (m != 1) {
    if (m & 1) {
      k = (k * n) % p;
    }
    n = (n * n) % p;
    m >>= 1;
  }
  return (n * k) % p;
}
int Judge(int x, int p) {
  int i;
  if (ModExp(x, p - 1, p) != 1) {
    return 0;
  }
  for (i = 1; i <= p - 2; ++i) {
    if (ModExp(x, i, p) == 1) {
      return 0;
    }
  }
  return 1;
}
int main() {
  int P, x, Sum;
  scanf("%d", &P);
  for (Sum = 0, x = 1; x < P; ++x) {
    Sum += Judge(x, P);
  }
  printf("%d\n", Sum);
  return 0;
}

```