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.util.*;

public class Dima_and_Lisa {

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int n = sc.nextInt();
        int max = n;

        while(true){
            boolean isPrime = true;

            for(int i = 2; i*i<=max; i++){
                if(max%i == 0){
                    isPrime = false;
                    break;
                }
            }
            if(isPrime)
                break;
            else
                max--;
        }

        int left = n - max;
        if(left == 0){
            System.out.println(1);
            System.out.println(max);
        }else if(left ==2){
            System.out.println(2);
            System.out.println(2+" "+max);
        }else{
            for(int i = 2; i<=left; i++){
                if(prime(i) && prime(left-i)){
                    System.out.println(3);
                    System.out.println(i+" "+max+" "+(left-i));
                    break;
                }
            }
        }
    }

    static boolean prime(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 sort(int[] a) {
        int n = a.length;
        for (int i = 0; i < n; i++) {
            int val = random.nextInt(n);
            int cur = a[i];
            a[i] = a[val];
            a[val] = cur;
        }
        Arrays.sort(a);
    }

    static void sortl(long[] a) {
        int n = a.length;
        for (int i = 0; i < n; i++) {
            int val = random.nextInt(n);
            long cur = a[i];
            a[i] = a[val];
            a[val] = cur;
        }
        Arrays.sort(a);
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

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

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

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

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

}