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.BufferedWriter;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.BitSet;

public class Main {
	public static void main(String[] args) throws NumberFormatException,
	IOException {
		Solve solve = new Solve();
		solve.solve();
	}
}
class Solve{
	void solve() throws NumberFormatException, IOException{
		ContestScanner in = new ContestScanner();
		Writer out = new Writer();
		int n = in.nextInt();
		int p = 0;
		for(int i=n; i>0; i--){
			if(isPrime(i)){
				p = i;
				break;
			}
		}
		n -= p;
		if(n==0){
			System.out.println(1);
			System.out.println(p);
			return;
		}
		BitSet prime = prime(10000000);
		for(int i=prime.nextSetBit(0); i!=-1; i=prime.nextSetBit(i+1)){
			if(n==i){
				System.out.println(2);
				System.out.println(p+" "+i);
				return;
			}
			if(prime.get(n-i)){
				System.out.println(3);
				System.out.println(p+" "+i+" "+(n-i));
				return;
			}
		}
	}
	
	public boolean isPrime(int n){
		for(int i=2; i*i<=n; i++){
			if(n%i==0) return false;
		}
		return true;
	}
	
	public static BitSet prime(int max){
		// ��������g���悤��
		BitSet prime = new BitSet(max+1);
		prime.set(2, max+1);
		for(int i=2; i!=-1; i=prime.nextSetBit(i+1)){
			for(int j=i*2; j<=max; j+=i){
				prime.clear(j);
			}
		}
		return prime;
	}
}

class Timer{
	long time;
	public void set(){time = System.currentTimeMillis();}
	public long stop(){return System.currentTimeMillis()-time;}
}
class Writer extends PrintWriter{
	public Writer(String filename) throws IOException
	{super(new BufferedWriter(new FileWriter(filename)));}
	public Writer() throws IOException{super(System.out);}
}
class ContestScanner {
	private BufferedReader reader;
	private String[] line;
	private int idx;
	public ContestScanner() throws FileNotFoundException 
	{reader = new BufferedReader(new InputStreamReader(System.in));}
	public ContestScanner(String filename) throws FileNotFoundException
	{reader = new BufferedReader(new InputStreamReader(new FileInputStream(filename)));}
	public String nextToken() throws IOException {
		if (line == null || line.length <= idx) {
			line = reader.readLine().trim().split(" ");
			idx = 0;
		}
		return line[idx++];
	}
	public String readLine() throws IOException{return reader.readLine();}
	public long nextLong() throws IOException, NumberFormatException
	{return Long.parseLong(nextToken());}
	public int nextInt() throws NumberFormatException, IOException
	{return (int) nextLong();}
	public double nextDouble() throws NumberFormatException, IOException 
	{return Double.parseDouble(nextToken());}
}