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

public class Main {
    static BigInteger steps = BigInteger.ZERO;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        BigInteger n = new BigInteger(in.next());
        solve(n);
    }

    static void solve(BigInteger i) {
        if (i.equals(BigInteger.ZERO)) {
            System.out.println(steps);
            return;
        }
        if (isPrime(i)) {
            steps = steps.add(BigInteger.ONE);
            System.out.println(steps);
            return;
        }
        for (int j = 1; i.compareTo(new BigInteger(String.valueOf(j))) >= 0; j++) {
            if (isPrime(new BigInteger(String.valueOf(j))) && i.remainder(new BigInteger(String.valueOf(j))).equals(BigInteger.ZERO)) {
                steps = steps.add(i.divide(new BigInteger(String.valueOf(j))));
                i = i.remainder(new BigInteger(String.valueOf(j)));
                break;
            }
        }
        solve(i);
    }

    static boolean isPrime(BigInteger i) {
        if (i.equals(new BigInteger("2"))) {
            return true;
        }
        if (i.equals(BigInteger.ONE)) {
            return false;
        }
        if (i.remainder(new BigInteger("2")).equals(BigInteger.ZERO)) {
            return false;
        }
        for (int j = 3; new BigInteger(String.valueOf(j)).pow(2).compareTo(i) <= 0; j++) {
            if (i.remainder(new BigInteger(String.valueOf(j))).equals(BigInteger.ZERO)) {
                return false;
            }
        }
        return true;
    }

}

