#include <bits/stdc++.h>
using namespace std;
bool isPrime(int n) {
  if (n < 2) return false;
  if (n == 2 || n == 3 || n == 5) return true;
  if (n == 4) return false;
  if (n > 5) {
    if (n % 2 == 0) return false;
    if (n % 3 == 0) return false;
    int t = sqrtl(n);
    for (int i = 5; i <= t; i += 6) {
      if (n % i == 0) return false;
    }
    for (int i = 7; i <= t; i += 6) {
      if (n % i == 0) return false;
    }
  }
  return true;
}
bool check2(int p, int n) {
  int re = 1;
  for (int i = 1; i <= p - 2; i++) {
    re = (re * n) % p;
    if (re % p == 1) return false;
  }
  re = (re * n) % p;
  return (re - 1) % p == 0;
}
int main() {
  int p;
  cin >> p;
  int count = 0;
  for (int i = 1; i < p; i++) {
    if (check2(p, i)) count++;
  }
  cout << count;
  return 0;
}

### Prompt

Generate 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>
using namespace std;
bool isPrime(int n) {
  if (n < 2) return false;
  if (n == 2 || n == 3 || n == 5) return true;
  if (n == 4) return false;
  if (n > 5) {
    if (n % 2 == 0) return false;
    if (n % 3 == 0) return false;
    int t = sqrtl(n);
    for (int i = 5; i <= t; i += 6) {
      if (n % i == 0) return false;
    }
    for (int i = 7; i <= t; i += 6) {
      if (n % i == 0) return false;
    }
  }
  return true;
}
bool check2(int p, int n) {
  int re = 1;
  for (int i = 1; i <= p - 2; i++) {
    re = (re * n) % p;
    if (re % p == 1) return false;
  }
  re = (re * n) % p;
  return (re - 1) % p == 0;
}
int main() {
  int p;
  cin >> p;
  int count = 0;
  for (int i = 1; i < p; i++) {
    if (check2(p, i)) count++;
  }
  cout << count;
  return 0;
}

```