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:
// Don't place your source in a package

import java.math.BigInteger;
import java.util.*;
import java.lang.*;
import java.io.*;



// Please name your class Main
public class Main {
    static FastScanner fs=new FastScanner();
    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();
        }
        int Int() {
            return Integer.parseInt(next());
        }

        long Long() {
            return Long.parseLong(next());
        }

        String Str(){
            return next();
        }
    }


    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(System.out);


        int T=1;
        for(int t=0;t<T;t++){
            int n=Int();
            Solution sol=new Solution(out);
            sol.solution(n);
        }
        out.flush();

    }


    public static int Int(){
        return fs.Int();
    }
    public static long Long(){
        return fs.Long();
    }
    public static String Str(){
        return fs.Str();
    }

}






class Solution{
    PrintWriter out;
    public Solution(PrintWriter out){
        this.out=out;
    }




    public void solution(int n){

        if(isp(n)){
            out.println(1);
            out.println(n);
            return;
        }

        if(isp(n-2)){
            out.println(2);
            out.println(2+" "+(n-2));
            return;
        }

        out.println(3);
        n-=3;
        for(int i=2;i<n;i++){
            if(isp(i)&&isp(n-i)){
                out.println(3+" "+i+" "+(n-i));
                return;
            }
        }

    }

    public boolean isp(int n){
        for(long i=2;i*i<=n;i++){
            if(n%i==0)return false;
        }
        return true;
    }













}



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                      free bug dog
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