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 int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const long double pi = acos(-1);
const int MOD = 1e9 + 7;
long long qnt = 0;
long long vai(long long x) {
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      if (i % 2) {
        return 1 + (x - i) / 2;
      } else {
        return x / 2;
      }
    }
  }
  return 1;
}
int main() {
  long long n;
  cin >> n;
  cout << vai(n) << endl;
  return 0;
}
