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;
char s[310000];
int n;
vector<long long> v;
int main() {
  cin >> n;
  for (long long i = 2; i <= 110000; i++) {
    bool b = 1;
    for (long long j = 2; j * j <= i; j++) {
      if (i % j == 0) {
        b = 0;
        break;
      }
    }
    if (b) v.push_back(i);
  }
  long long now = 1;
  for (int i = 0; i < v.size(); i++) {
    if (n % v[i] == 0) {
      n -= v[i];
      break;
    }
  }
  now += n / 2;
  cout << now;
  return 0;
}
