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;
long long n;
int a[100005] = {0}, q[100005], t = 0;
void init() {
  cin >> n;
  int m = sqrt(n);
  for (int i = 2; i <= m; i++) {
    if (a[i] == 0) q[++t] = i;
    for (int j = 1; j <= t && i * q[j] <= m; j++) {
      a[i * q[j]] = 1;
      if (i % q[j] == 0) break;
    }
  }
}
int main() {
  init();
  long long ans = 0;
  for (int i = 1; i <= t; i++) {
    if (n % q[i] == 0) {
      ans = n / q[i];
      break;
    }
  }
  if (ans == 0) ans++;
  cout << ans;
  return 0;
}
