Problem: Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.

More formally, you are given an odd numer n. Find a set of numbers pi (1 ≤ i ≤ k), such that

  1. 1 ≤ k ≤ 3
  2. pi is a prime
  3. <image>



The numbers pi do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.

Input

The single line contains an odd number n (3 ≤ n < 109).

Output

In the first line print k (1 ≤ k ≤ 3), showing how many numbers are in the representation you found.

In the second line print numbers pi in any order. If there are multiple possible solutions, you can print any of them.

Examples

Input

27


Output

3
5 11 11

Note

A prime is an integer strictly larger than one that is divisible only by one and by itself.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;

public class Solution{
	
	
	
	
    public static void main(String[] args) throws IOException {
     
    	FastScanner fs = new FastScanner();
    	PrintWriter out = new PrintWriter(System.out);
    	
    		
    		
    	int tt = 1;
    	while(tt-->0) {
    		
    		int n = fs.nextInt();
    		ArrayList<Integer> ans = new ArrayList<Integer>();
    		if(isPrime(n)) {
    			ans.add(n);
    		}
    		else {
    			//distance between consecutive primes is less.
    			for(int i=n-2;;i--) {
    				if(isPrime(i)) {
    					ans.add(i);
    					n -= i;
    					break;
    				}
    			}
    			if(isPrime(n)) {
    				ans.add(n);
    			}
    			else {
    				//goldbach's conjecture: every even number greater than 2 can be represented as the sum of two primes.
    				for(int i=2;i<n-1;i++) {
    					if(isPrime(i) && isPrime(n-i)) {
    						ans.add(i);
    						ans.add(n-i);
    						break;
    					}
    				}
    			}
    		}
    		
    		out.println(ans.size());
    		for(int a: ans) out.print(a+" ");
    		out.println();
    		
    		
    	}
    	
    		
    		
    		
    	out.close();
    		
    }
    
    
    
    //O(sqrt(n))
    static boolean isPrime(int n) {
    	for(int i=2;i*i<=n;i++) {
    		if(n%i==0) return false;
    	}
    	return true;
    }
   
    
    
    
    
    
    
    static final Random random=new Random();
    	
    static void ruffleSort(int[] a) {
    	int n=a.length;//shuffle, then sort 
    	for (int i=0; i<n; i++) {
    		int oi=random.nextInt(n); int temp=a[oi];
    		a[oi]=a[i]; a[i]=temp;
    	}
    	Arrays.sort(a);
    }
   
  
    
    	
    	
    static class FastScanner{
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer st = new StringTokenizer("");
     
    	public String next(){
    		while(!st.hasMoreElements()){
    			try{
    				st = new StringTokenizer(br.readLine());
    			} catch(IOException e){
    				e.printStackTrace();
    			}
    		}
    		return st.nextToken();
    	}
    		
    	public String nextLine() throws IOException {
    		return br.readLine();
    	}
    		
    	public int nextInt(){
    		return Integer.parseInt(next());
    	}
     
    	public int[] readArray(int n){
    		int[] a = new int[n];
    		for(int i=0;i<n;i++)
    			a[i] = nextInt();
    		return a;
    	}
    		
    	public long nextLong() {
    		return Long.parseLong(next());
    	}
    		
    	public char nextChar() {
    		return next().toCharArray()[0];
    	}
    }
   	
}