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 long long N = 2e5 + 9;
int main() {
  ios_base::sync_with_stdio(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, st, en, d;
    cin >> n >> st >> en >> d;
    if (abs(st - en) % d == 0) {
      cout << abs(st - en) / d;
      return 0;
    }
    long long ans = 1e18;
    if ((en - 1) % d == 0) ans = min(ans, (st - 1 + d - 1) / d + (en - 1) / d);
    if ((n - en) % d == 0) ans = min(ans, (n - st + d - 1) / d + (n - en) / d);
    cout << (ans == 1e18 ? -1 : ans) << "\n";
  }
}
