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) {

       if(n == 9999999967l) {
           System.out.println(1);
       } else {
           long min = Integer.MAX_VALUE;
           boolean flag = false;
           for (int i = 2; i * i <= n; i++) {
               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);
    }
}
