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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Divisor_Sub {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next(){
            while(st==null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        String nextLine(){
            String str = "";
            try
            {
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }
            return str;
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
    }
    public static void main(String[] args) {
       FastReader in = new FastReader();
       long n = in.nextLong();
       long ans = 0;
       while(n % 2 != 0){
           n -= get(n);
           ans++;
       }
       ans+=(n/2);
       System.out.println(ans);
    }
   static long get(long n){
        for(long i = 2; i * i <= n; i++){
            if(n % i == 0){
                return i;
            }
        }
        return n;
   }
}
