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.util.Scanner;

public class ProblemB {
	static boolean[] isPrime;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		sc.close();
		isPrime = new boolean[n + 1];
		aryPrime();
		int cnt = 0;
		while(n != 0) {
			if(n % 2 == 0) {
				n /= 2;
				cnt++;
				n -= 2;
			}else {
				if(isPrime[n]) {
					cnt++;
					break;
				}else {
					for(int i = 3; i * i < n; i += 2) {
						if(n % i == 0) {
							n /= i;
							cnt++;
							n -= i;
						}
					}
				}
			}
		}
		System.out.println(cnt);
	}
	static void aryPrime(){
        int l = isPrime.length;
        for(int i = 0; i < l; i++) {
            isPrime[i] = true;
        }
        isPrime[0] = false;
        isPrime[1] = false;
        for(int i = 2; i <= (int)Math.sqrt(l); i++) {
            if(isPrime[i]) {
                for(int j = i * 2; j < l; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }
}