Problem: Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.

Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k. 

Input

The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).

Output

The first line of the output contains a single integer k — maximum possible number of primes in representation.

The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.

Examples

Input

5


Output

2
2 3


Input

6


Output

3
2 2 2

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main {

	void solve (FastScanner in, PrintWriter out) {
		
		int n = in.nextInt();
		ArrayList<Integer> list = new ArrayList<>();
		
		if (n%2 == 0) {
			for (int i=0; i<n/2; i++) list.add(2);
		}
		else {
			for (int i=0; i<n/2-1; i++) list.add(2);
			list.add(3);
		}
		
		out.println(list.size());
		for (int i=0; i<list.size(); i++) {
			out.print((i==0?"":" ")+list.get(i));
		}
		out.println();
		
	}

	void print (Object... ar) {
		System.out.println(Arrays.deepToString(ar));
	}

	public static void main(String[] args) {
		FastScanner in = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		Main main = new Main();
		main.solve(in, out);
		in.close();
		out.close();
	}

	static class FastScanner {

		private InputStream in;
		private byte[] buffer = new byte[1024];
		private  int length = 0, p = 0;

		public FastScanner (InputStream stream) {
			in = stream;
		}
		public boolean hasNextByte () {
			if (p < length) return true;
			else {
				p = 0;
				try {length = in.read(buffer);}
				catch (Exception e) {e.printStackTrace();}
				if (length <= 0) return false;
			}
			return true;
		}
		public int readByte () {
			if (hasNextByte() == true) return buffer[p++];
			return -1;
		}
		public boolean isPrintable (int n) {return 33<=n&&n<=126;}
		public void skip () {
			while (hasNextByte() && !isPrintable(buffer[p])) p++;
		}
		public boolean hasNext () {skip(); return hasNextByte();}
		public String next () {
			if (!hasNext()) throw new NoSuchElementException();
			StringBuilder sb = new StringBuilder();
			int t = readByte();
			while (isPrintable(t)) {
				sb.appendCodePoint(t);
				t = readByte();
			}
			return sb.toString();
		}
		public String[] nextArray (int n) {
			String[] ar = new String[n];
			for (int i=0; i<n; i++) ar[i] = next();
			return ar;
		}
		public int nextInt () {return Math.toIntExact(nextLong());}
		public int[] nextIntArray (int n) {
			int[] ar = new int[n];
			for (int i=0; i<n; i++) ar[i] = nextInt();
			return ar;
		}
		public long nextLong () {
			if (!hasNext()) throw new NoSuchElementException();
			boolean minus = false;
			int temp = readByte();
			if (temp == '-') {
				minus = true;
				temp = readByte();
			}
			if (temp<'0' || '9'<temp) throw new NumberFormatException();
			long n = 0;
			while (isPrintable(temp)) {
				if ('0'<=temp && temp<='9') {
					n *= 10;
					n += temp - '0';
				}
				else throw new NumberFormatException();
				temp = readByte();
			}
			return minus? -n : n;
		}
		public double nextDouble () {
			return Double.parseDouble(next());
		}
		public void close () {
			try {in.close();}
			catch(Exception e){}
		}
	}

}