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.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;

public class Main{
    public static void main(String[] args)throws IOException{
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	PrintWriter out = new PrintWriter(System.out);
	long n = Long.parseLong(br.readLine());
	BigInteger aux = new BigInteger(""+n);
	// byte sieve[] = new byte [100001];
	// int primos[] = new int[317];
	// sieve[0] = 1;
	// sieve[1] = 1;
	// int cont = 0;
	// for(int i = 2; i<=10000;i++){
	//     if(sieve[i] == 0 && i*i<=100000){
	// 	primos[cont] = i;
	// 	cont++;
	// 	for(int j = i*i;j<=100000;j+=i){
	// 	    sieve[j] = 1;
	// 	}
	//     }
	// }
	if(aux.isProbablePrime(20)){
	    out.println(1);
	}else{
	    long ans = 0;
	    long k = 2;
	    while(true){
		aux = new BigInteger(""+n);
		if(aux.isProbablePrime(20)){
		    ans = 1;
		    break;
		}else{
		    while(true){
			aux = new BigInteger(""+k);
			if(n%k == 0){
			    if(aux.isProbablePrime(20)){
				ans = n/k;
				break;
			    }
			}else{
			    k++;
			}
		    }
		    break;
		}
	    }
	    out.println(ans);
	}
	
	out.close();
    }
}
