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;
int vis[100000];
int a[100010], cnt;
void Init() {
  memset(vis, 0, sizeof(vis));
  int i, j;
  for (i = 2; i < 100000; i++) {
    if (!vis[i]) {
      a[cnt++] = i;
      for (j = i + i; j < 100000; j += i) {
        vis[j] = 1;
      }
    }
  }
}
int main() {
  Init();
  long long n, ans = 0;
  cin >> n;
  for (int i = 0; i < cnt; i++)
    if (n % a[i] == 0 && (n - a[i]) % 2 == 0) {
      cout << (n - a[i]) / 2 + 1 << endl;
      return 0;
    } else if (n % a[i] == 0) {
      cout << n / a[i] << endl;
      return 0;
    }
  cout << 1 << endl;
  return 0;
}
