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.
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;

public class B {

    public static void main(String []arg) {
        Scanner sc = new Scanner(System.in);
         long n = sc.nextLong();
         if(isPrime(n)){
             System.out.println(1);

         }
         else if (n%2==0){
             System.out.println(n/2);
         }
         else {
             for(long i = 3;true;i+=2){
                 if(n%i==0){
                     if(isPrime(i)){
                         System.out.println(n/i);
                         break;
                     }
                 }
             }
         }

    }
    static  boolean isPrime( long n)
    {
        if (n < 2)  return false;
        if (n < 4)  return true;

        if (n%2 == 0 || n%3 == 0) return false;

        for (long i = 5 ; i * i <= n ; i = i + 6 )
            if ( n % i == 0 || n % ( i + 2 ) == 0)
                return false ;

        return true;
    }
    static long gcd(long a, long b)
    {
        return b == 0 ? (a < 0 ? -a : a) : gcd(b, a % b);
    }
    static long lcm(long a, long b)
    {
        long lcm = (a / gcd(a, b)) * b;
        return lcm > 0 ? lcm : -lcm ;
    }

}