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.*;
import java.math.*;
import java.util.*;


public class Program {

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;

    static final int MOD = 1000000007;
    static final double INF = 10000000000.0;
    
    boolean isPrime(int n)
    {
    	if (n == 1)
    		return false;
    	for(int i = 2; i*i <= n; i++)
    		if (n % i == 0)
    			return false;
    	return true;
    }
    
 	void solve() throws IOException {
 		int n = nextInt();
 		if (isPrime(n))
 		{
 			out.println(1);
 			out.println(n);
 		}
 		else if (n % 2 == 0)
 		{
 			for(int i = 1; i <= n; i+=2)
 			if (isPrime(i) && isPrime(n-i))
 			{
 	 			out.println(2);
 				out.println(i + " " + (n-1));
 				break;
 			}
 		}
 		else
 		{
 			for(int i = 1; i <= n; i+=2)
 			if (isPrime(i) && isPrime(n-i-3))
 			{
 				out.println(3);
 				out.println(i + " " + (n-i-3) + " " + 3);
 				break;
 			}
 		}
    }

    Program() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        new Program();
    }

    String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }

    String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
    
    class Pair implements Comparable<Pair>{

        char first;
        int second;

        public Pair(char an1, int an2){
            first = an1;
            second = an2;
        }

        @Override
        public int compareTo(Pair o) {
            if (first - o.first == 0)
            	return second - o.second;
            else
            	return first - o.first;
        }
    }
}