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;
long long l(long long v) {
  for (long long r = 2; r * r <= v; r++) {
    if (v % r == 0) return r;
  }
  return v;
}
int main() {
  long long b;
  cin >> b;
  int ans = 0;
  while (b % 2) {
    ans++;
    b -= l(b);
  }
  cout << ans + b / 2;
  return 0;
}
