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>
long long cnt = 0;
void alg(int n) {
  if (n <= 1)
    return;
  else {
    int tmp = sqrt((double)n) + 1;
    int divisor = 0, sub;
    if (n % 2 == 0 && 2 != n) {
      divisor = 2;
    } else {
      for (int i = 3; i <= tmp; i += 2) {
        if (n % i == 0) {
          divisor = i;
          break;
        }
      }
    }
    if (divisor)
      sub = n - divisor;
    else
      sub = 0;
    cnt++;
    alg(sub);
    return;
  }
}
int main() {
  long long num;
  scanf("%d", &num);
  alg(num);
  printf("%d\n", cnt);
  return 0;
}
