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 N = 200005;
const long long inf = 0x3f3f3f3fll;
const long long mod = 998244353ll;
int n, m;
int main() {
  ios::sync_with_stdio(0);
  int i, j;
  long long x;
  cin >> x;
  if (!(x & 1))
    cout << x / 2;
  else {
    long long k;
    for (k = 2; k * k <= x; k++) {
      if (x % k == 0) break;
    }
    if (k * k > x)
      cout << 1;
    else
      cout << (x - k) / 2 + 1;
  }
  return 0;
}
