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 maxn = 100007;
const long long mod = 998244353;
const long long ool = 1e18 + 7;
const int ooi = 1e9 + 7;
const int apsz = 26;
const int o = 1;
const int maxd = 20;
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  long long tn = n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i) continue;
    tn -= i;
    break;
  }
  if (tn == n) return cout << 1 << endl, 0;
  long long ans = 1;
  cout << ans + tn / 2 << endl;
  return 0;
}
