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.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.BitSet;
import java.util.NoSuchElementException;
import java.util.Scanner;

public class Main {
    //
    // private void test() {
    // Random random = new Random();
    // for (int i = 0; i < 100; i++) {
    // int next = random.nextInt((int) 1e9);
    // if (next % 2 == 0) {
    // i--;
    // continue;
    // }
    // System.out.println(next);
    // solve(next);
    // }
    // System.out.println();
    //
    // }

    public void solve(int n) {
	int m = (int) Math.sqrt(n) + 1;
	ArrayList<Integer> primes = sieveEratos(m);

	{// k=1
	    boolean one = true;
	    for (int p : primes) {
		if (n % p == 0) {
		    one = false;
		    break;
		}
	    }
	    if (one) {
		System.out.println(1);
		System.out.println(n);
		return;
	    }
	}
	{
	    // k=2
	    boolean two = true;
	    int rest = n - 2;
	    for (int p : primes) {
		if (rest % p == 0) {
		    two = false;
		    break;
		}
	    }
	    if (two) {
		System.out.println(2);
		System.out.println(rest + " " + 2);
		return;
	    }
	}

	// guaranteed
	System.out.println(3);
	for (int i = 0; i < primes.size(); i++) {
	    int p1 = primes.get(i);
	    for (int j = 0; j < primes.size(); j++) {
		int p2 = primes.get(j);
		int rest = n - p1 - p2;
		for (int k = 0; k < primes.size(); k++) {
		    if (rest % primes.get(k) == 0) {
			break;
		    }
		    if (k == primes.size() - 1) {
			System.out.println(p1 + " " + p2 + " " + rest);
			return;
		    }
		}

	    }
	}

    }

    private ArrayList<Integer> sieveEratos(int n) {
	BitSet primeset = new BitSet();
	primeset.flip(2, n + 1);
	for (int q = primeset.nextSetBit(2); q != -1 && q * q <= n; q = primeset
		.nextSetBit(q + 1)) {
	    for (int i = q * q; i <= n; i += q) {
		primeset.clear(i);
	    }
	}

	ArrayList<Integer> primes = new ArrayList<>(primeset.cardinality());
	for (int q = primeset.nextSetBit(2); q != -1; q = primeset
		.nextSetBit(q + 1)) {
	    primes.add(q);
	}
	return primes;
    }

    public static void main(String[] args) {
	Scanner scanner = new Scanner(System.in);
	int n = scanner.nextInt();
	scanner.close();
	new Main().solve(n);
    }
}

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
	if (ptr < buflen) {
	    return true;
	} else {
	    ptr = 0;
	    try {
		buflen = in.read(buffer);
	    } catch (IOException e) {
		e.printStackTrace();
	    }
	    if (buflen <= 0) {
		return false;
	    }
	}
	return true;
    }

    private int readByte() {
	if (hasNextByte())
	    return buffer[ptr++];
	else
	    return -1;
    }

    private static boolean isPrintableChar(int c) {
	return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
	while (hasNextByte() && !isPrintableChar(buffer[ptr]))
	    ptr++;
    }

    public boolean hasNext() {
	skipUnprintable();
	return hasNextByte();
    }

    public String next() {
	if (!hasNext())
	    throw new NoSuchElementException();
	StringBuilder sb = new StringBuilder();
	int b = readByte();
	while (isPrintableChar(b)) {
	    sb.appendCodePoint(b);
	    b = readByte();
	}
	return sb.toString();
    }

    public int nextInt() {
	return (int) nextLong();
    }

    public long nextLong() {
	if (!hasNext())
	    throw new NoSuchElementException();
	long n = 0;
	boolean minus = false;
	int b = readByte();
	if (b == '-') {
	    minus = true;
	    b = readByte();
	}
	if (b < '0' || '9' < b) {
	    throw new NumberFormatException();
	}
	while (true) {
	    if ('0' <= b && b <= '9') {
		n *= 10;
		n += b - '0';
	    } else if (b == -1 || !isPrintableChar(b)) {
		return minus ? -n : n;
	    } else {
		throw new NumberFormatException();
	    }
	    b = readByte();
	}
    }
}