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 DivisorSubtraction {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long n = in.nextLong();
        System.out.println(cnt(n));
    }

    static long cnt(long n) {
        if(n % 2 == 0) return n/2;
        if(isPrime(n)) return 1;
        for(long i = 3; i < n; i+=2) {
            if(isPrime(i)) {
                if(n % i == 0) return 1 + cnt(n-i);
            }
        }
        return 0;
    }

    static boolean isPrime(long n) {
        if(n < 2) return false;
        if(n == 2 || n == 3) return true;
        if(n%2 == 0 || n%3 == 0) return false;
        long sqrtN = (long)Math.sqrt(n)+1;
        for(long i = 6L; i <= sqrtN; i += 6) {
            if(n%(i-1) == 0 || n%(i+1) == 0) return false;
        }
        return true;
    }
}
