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 criba[100000 + 1];
int main() {
  long long n, j;
  int i, p;
  scanf("%I64d", &n);
  memset(criba, 0, sizeof criba);
  for (i = 2; i <= 100000; ++i)
    if (!criba[i]) {
      if (n % i == 0) {
        printf("%I64d", n / i);
        break;
      }
      j = i * 1LL * i;
      while (j <= 100000) {
        criba[j] = true;
        j += i;
      }
    }
  return 0;
}
