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;
const int MAXN = (int)1e5 + 5;
const int MOD = (int)1e9 + 7;
int main() {
  long long n, m = 2;
  cin >> n;
  for (m = 2; m * m <= n; m++) {
    if (n % m == 0) break;
  }
  if (n % m != 0) m = n;
  if (n == m)
    printf("1\n");
  else {
    if (m == 2)
      printf("%lld\n", n / m);
    else
      printf("%lld\n", (n - m) / 2 + 1);
  }
  return 0;
}
