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.*;
import java.math.*;
 
public class Main {
 
    FastScanner in;
    PrintWriter out;
 
    static class FastScanner {
 
        BufferedReader br;
        StringTokenizer st;
 
        FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = null;
        }
 
        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
 
        String nextLine() throws IOException {
            if (st == null || !st.hasMoreTokens())
                return br.readLine();
            StringBuilder result = new StringBuilder(st.nextToken());
            while (st.hasMoreTokens()) {
                result.append(" ");
                result.append(st.nextToken());
            }
            return result.toString();
        }
 
        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
 
        long nextLong() throws IOException {
            return Long.parseLong(next());
        }
 
        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
 
    }
 
    void run() throws IOException {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out, false);
        solve();
        out.close();
    }
 
    public static void main(String[] args) throws IOException{
        new Main().run();
    }
 
    public void printArr(int[] arr){
        for(int i = 0; i < arr.length; i++){
            out.print(arr[i] + " ");
        }
        out.println();
    }
 
    public long gcd(long a, long b){
        if(a == 0) return b;
        return gcd(b % a, a);
    }

    public boolean isPrime(long num){

        if(num == 0 || num == 1){
            return false;
        }

        for(int i = 2; i * i <= num; i++){
            if(num % i == 0){
                return false;
            }
        }

        return true;
    }


    public void solve() throws IOException{

        int n = in.nextInt();

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

        int p0 = -1, p1 = -1, p2 = -1;
        for(int i = n - 2; i >= 0; i--){
            if(isPrime(i)){
                p0 = i;
                break;
            }
        }

        if(isPrime(n - p0)){
            out.println(2);
            out.println(p0 + " " + (n - p0));
            return;
        }

        for(int i = 2; i <= n - p0; i++){

            if(isPrime(i) && isPrime(n - p0 - i)){
                p1 = i;
                p2 = n - p0 - i;
            }
        }

        out.println(3);
        out.println(p0 + " " + p1 + " " + p2);
        return;
    }

    

}