Problem: Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.

Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k. 

Input

The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).

Output

The first line of the output contains a single integer k — maximum possible number of primes in representation.

The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.

Examples

Input

5


Output

2
2 3


Input

6


Output

3
2 2 2

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class Tester {
public static void main(String[] args){
	Scanner s =new Scanner(System.in);
	int n=s.nextInt();
	int[] arr=new int[n];
	arr=SeiveOfEratosthenes(n);
	int count=0;
	int rem=n;
	ArrayList<Integer> a=new ArrayList<Integer>();
	for(int i=0;i<arr.length;i++){
		int temp=arr[i];
		while(true){
			if(rem==0)
				break;
			else if(rem-temp>=temp || rem==temp){
					rem=rem-temp;
					a.add(temp);
					count++;
			}
			else 
			break;
	}
		if(temp==0)
			break;
	}
	System.out.println(count);
	for(int i=0;i<a.size();i++)
	System.out.print(a.get(i)+" ");
	
}

static int[] SeiveOfEratosthenes(int n){
	boolean prime[]=new boolean[n+1];
	for(int i=0;i<=n;i++){
		prime[i]=true;
	}
	for(int p=2;p*p<=n;p++){
		if(prime[p]==true){
			for(int i=p*2;i<=n;i=i+p){
				prime[i]=false;
			}
		}
	}
	int[] arr=new int[n];
	int j=0;
	for(int i=2;i<=n;i++)
	{
		if(prime[i]==true){
			arr[j]=i;
			j++;
		}
	}
	
	return arr;
	
}
}