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.
//package codeforces.div2;

import java.math.BigInteger;
import java.util.Scanner;

public class DivisorSubtraction {

   static public void print(long n) {


           long min = Integer.MAX_VALUE;
           boolean flag = false;
           if(n%2 == 0) {
               System.out.println(1 + (n - 2) / 2);

           } else {
               for (long i = 3; i * i <= n; i = i + 1) {
                   if (n % i == 0) {
                       min = Math.min(min, i);
                       //min = Math.min(min, n / i);
                       flag = true;
                       break;
                   }
                   if (flag) {
                       break;
                   }
               }
               if (min == Integer.MAX_VALUE) {
                   System.out.println(1);
               } else {
                   System.out.println(1 + (n - min) / 2);
               }
           }

    }

    public static void main(String []args) {
        Scanner sc = new Scanner(System.in);

        long x = sc.nextLong();
        print(x);
    }
}
