#include <bits/stdc++.h>
using namespace std;
int p;
int xpow(int num, int pow) {
  if (pow == 1) return (num) % p;
  if (pow == 0) return 1;
  int z = xpow(num, pow / 2);
  if (pow % 2)
    return ((num) % p * ((z) % p * (z) % p) % p) % p;
  else
    return (z % p * z % p) % p;
}
bool is_prim(int x) {
  if (xpow(x, p - 1) % p != 1) return 0;
  int z = 2;
  while (xpow(x, p - z) != 1 && z <= p - 1) z++;
  if (z == p) return 1;
  return 0;
}
int main() {
  cin >> p;
  int i = p - 1;
  int count = 0;
  while (i > 0) {
    if (is_prim(i)) count++;
    i--;
  }
  cout << count << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 p;
int xpow(int num, int pow) {
  if (pow == 1) return (num) % p;
  if (pow == 0) return 1;
  int z = xpow(num, pow / 2);
  if (pow % 2)
    return ((num) % p * ((z) % p * (z) % p) % p) % p;
  else
    return (z % p * z % p) % p;
}
bool is_prim(int x) {
  if (xpow(x, p - 1) % p != 1) return 0;
  int z = 2;
  while (xpow(x, p - z) != 1 && z <= p - 1) z++;
  if (z == p) return 1;
  return 0;
}
int main() {
  cin >> p;
  int i = p - 1;
  int count = 0;
  while (i > 0) {
    if (is_prim(i)) count++;
    i--;
  }
  cout << count << '\n';
  return 0;
}

```