You are given an integer number n. The following algorithm is applied to it:

  1. if n = 0, then end algorithm; 
  2. find the smallest prime divisor d of n; 
  3. subtract d from n and go to step 1. 



Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer n (2 ≤ n ≤ 10^{10}).

Output

Print a single integer — the number of subtractions the algorithm will make.

Examples

Input


5


Output


1


Input


4


Output


2

Note

In the first example 5 is the smallest prime divisor, thus it gets subtracted right away to make a 0.

In the second example 2 is the smallest prime divisor at both steps.
#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long p) {
  for (int(i) = 2; (i) <= (sqrt(p)); (i)++) {
    if (p % i == 0) return false;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  long long int N;
  long long p = 2;
  cin >> N;
  while (N) {
    if (p > 2 && p % 2 == 0) {
      p++;
      continue;
    }
    if (isPrime(p)) {
      if (N % p == 0) {
        cout << N / p << endl;
        break;
      }
    }
    p++;
  }
  return 0;
}
