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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int N = in.nextInt();

            ArrayList<Long> primes = primes(100000);

            if (prime(N)) {
                out.println("1");
                out.println(N);
                return;
            }

            long cand1 = N - 2;
            if (cand1 >= 0 && prime(cand1)) {
                out.println(2);
                out.println("2 " + cand1);
                return;
            }

            for (int i = 0; i < primes.size(); i++) {
                for (int j = i; j < primes.size(); j++) {
                    long cand = (long) N - primes.get(i) - primes.get(j);
                    if (cand >= 0 && prime(cand)) {
                        out.println("3");
                        out.println(primes.get(i) + " " + primes.get(j) + " " + cand);
                        return;
                    }
                }
            }
        }

        private boolean prime(long n) {
            for (int i = 2; i * i <= n; i++) {
                if (n % i == 0) return false;
            }
            return true;
        }

        private ArrayList<Long> primes(int MAX) {
            ArrayList<Long> primes = new ArrayList<>(1000);
            boolean compose[] = new boolean[MAX + 1];
            for (int i = 2; i < MAX; i++) {
                if (!compose[i]) {
                    primes.add((long) i);
                    for (int j = 2 * i; j < MAX; j += i) {
                        compose[j] = true;
                    }
                }
            }
            return primes;
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

    }
}