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:
/**
 * Created by Ahmad on 1/3/2017.
 */

import java.io.*;
import java.util.*;

public class C {
    public static void main(String[] args) throws Exception {
        FastScannerC fs = new FastScannerC();
        PrintWriter out = new PrintWriter(System.out);

        int n = fs.nextInt();
        if(isPrime(n)) {
            out.println("1\n" + n);
        } else {
            int N = n-2;
            while(!isPrime(N)) N--;
            //first prime = N
            int nn = n - N;
            for(int i = 2; i <= nn; i++) {
                if(isPrime(i) && isPrime(nn - i)) {
                    if(nn-i != 0 && nn-i != 1) out.printf("%d\n%d %d %d\n", 3, N, i, nn-i);
                    else if(nn-i != 1) out.printf("%d\n%d %d\n", 2, N, i);
                    else continue;
                    break;
                }
            }
        }
        out.close();

    }

    static boolean isPrime(int n) {
        for(int i = 2; i <= Math.sqrt(n); i++) {
            if(n%i == 0) return false;
        }
        return true;
    }
}

class FastScannerC {
    BufferedReader br;
    StringTokenizer st;

    public FastScannerC() {
        try	{
            br = new BufferedReader(new InputStreamReader(System.in));
            st = new StringTokenizer(br.readLine());
        } catch (Exception e){e.printStackTrace();}
    }

    public String next() {
        if (st.hasMoreTokens())	return st.nextToken();
        try {st = new StringTokenizer(br.readLine());}
        catch (Exception e) {e.printStackTrace();}
        return st.nextToken();
    }

    public int nextInt() {return Integer.parseInt(next());}

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

    public double nextDouble() {return Double.parseDouble(next());}

    public String nextLine() {
        String line = "";
        try {line = br.readLine();}
        catch (Exception e) {e.printStackTrace();}
        return line;
    }

    public Integer[] nextIntegerArray(int n) {
        Integer[] a = new Integer[n];
        for(int i = 0; i < n; i++) a[i] = nextInt();
        return a;
    }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for(int i = 0; i < n; i++) a[i] = nextInt();
        return a;
    }

    public char[] nextCharArray() {
        return next().toCharArray();
    }
}