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 mod = 1e9 + 7;
const double eps = 1e-9;
const int maxn = 3e5 + 5;
int main() {
  long long n;
  scanf("%lld", &n);
  if ((n & 1) == 0) {
    printf("%lld\n", n / 2);
    return 0;
  }
  long long ans = 1;
  long long sn = sqrt(n), p = -1;
  for (long long i = 3; i <= sn; ++i) {
    if (n % i) continue;
    p = i;
    break;
  }
  if (p == -1) p = n;
  n -= p;
  ans += n / 2;
  printf("%lld\n", ans);
  return 0;
}
