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.math.BigInteger;
import java.util.*;

public class cf584d implements Runnable {
	public static final String taskname = cf584d.class.getName();

	public StringTokenizer strtok;
	public BufferedReader inr;
	public PrintWriter out;

	public static void main(String[] args) {
		new Thread(new cf584d()).start();
	}

	public void run() {
		Locale.setDefault(Locale.US);
		boolean oj = System.getProperty("ONLINE_JUDGE") != null;
		try {
			inr = new BufferedReader(oj ? new InputStreamReader(System.in, "ISO-8859-1") : new FileReader(taskname + ".in"));
			out = new PrintWriter(oj ? new OutputStreamWriter(System.out, "ISO-8859-1") : new FileWriter(taskname + ".out"));
			solve();
		} catch (Exception e) {
			if (!oj)
				e.printStackTrace();
			System.exit(9000);
		} finally {
			out.flush();
			out.close();
		}
	}

	public String nextToken() throws IOException {
		while (strtok == null || !strtok.hasMoreTokens()) {
			strtok = new StringTokenizer(inr.readLine());
		}
		return strtok.nextToken();
	}

	public int nextInt() throws NumberFormatException, IOException {
		return Integer.parseInt(nextToken());
	}

	public long nextLong() throws NumberFormatException, IOException {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws NumberFormatException, IOException {
		return Double.parseDouble(nextToken());
	}

	// Solution

	boolean isPrime(int i) {
		for (int d = 2; d <= Math.sqrt(i) + 1; d++) {
			if (i % d == 0) {
				return false;
			}
		}
		return true;
	}

	public void solve() throws NumberFormatException, IOException {
		ArrayList<Integer> primes = new ArrayList<Integer>();
		primes.add(2);
		for (int i = 2; i < 5000; i++) {
			if (isPrime(i)) {
				primes.add(i);
			}
		}

		HashSet<Integer> pregen = new HashSet<Integer>(primes);
		for (int i : primes) {
			for (int j : primes) {
				pregen.add(i + j);
			}
		}

		int n =  nextInt();
		if (Collections.binarySearch(primes, n - 4) >= 0) {
			out.printf("3\n2 2 %d", n - 4);
			return;
		}

		if ((new BigInteger(((Integer)(n)).toString())).isProbablePrime(15)) {
			out.printf("1\n%d", n);
			return;
		}

		for (int v : pregen) {
			if (n - v < 2) {
				continue;
			}
			BigInteger nnn = new BigInteger(((Integer)(n - v)).toString());
			if (nnn.isProbablePrime(15)) {
				// reconstructing
				if (Collections.binarySearch(primes, v) >= 0) {
					out.printf("2\n%d %d", v, n - v);
					return;
				}
				else {
					for (int vv : primes) {
						if (Collections.binarySearch(primes, v - vv) >= 0) {
							out.printf("3\n%d %d %d", vv, v - vv, n - v);
							return;
						}
					}
				}
			}


		}
	}
}