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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;

/*
 * Target : Tshirt in a major programming competition.
 */
public class Main implements Runnable {

	int mod = 998244353;
	int mod2 = mod;
	int sz[];
	int id[];
	int tree[];

	void test() throws IOException {
		Random rnd = new Random();
	}

	private void solve() throws IOException {
		int n = nextInt();
		assert (n % 2 == 1);
		boolean sieve[] = new boolean[(int) 1e5 + 1];
		Arrays.fill(sieve, true);
		sieve[0] = false;
		sieve[1] = false;
		for (int i = 2; i * i < sieve.length; ++i) {
			if (sieve[i])
				for (int j = i * i; j < sieve.length; j += i)
					sieve[j] = false;
		}
		if (isPrime(n)) {
			pw.println(1);
			pw.println(n);
		} else {
			pw.println(3);
			int len = (int) 1e5;
			for (int i = 0; i <= len; ++i) {
				if (sieve[i]) {
					if (isPrime(n - 2 * i)) {
						pw.println(i + " " + i + " " + (n - 2 * i));
						return;
					}
				}
			}
		}

	}

	boolean isPrime(int n) {
		for (int i = 3; i * i <= n; i += 2) {
			if (n % i == 0)
				return false;
		}
		return true;
	}

	BufferedReader br;
	StringTokenizer st;
	PrintWriter pw;

	public static void main(String args[]) {
		new Main().run();
	}

	public void run() {
		try {
			br = new BufferedReader(new InputStreamReader(System.in));
			pw = new PrintWriter(System.out);
			st = null;
			solve();
			pw.flush();
			pw.close();
			br.close();
		} catch (IOException e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

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

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

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

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