Given 2 integers u and v, find the shortest array such that [bitwise-xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of its elements is u, and the sum of its elements is v.

Input

The only line contains 2 integers u and v (0 ≤ u,v ≤ 10^{18}).

Output

If there's no array that satisfies the condition, print "-1". Otherwise:

The first line should contain one integer, n, representing the length of the desired array. The next line should contain n positive integers, the array itself. If there are multiple possible answers, print any.

Examples

Input


2 4


Output


2
3 1

Input


1 3


Output


3
1 1 1

Input


8 5


Output


-1

Input


0 0


Output


0

Note

In the first sample, 3⊕ 1 = 2 and 3 + 1 = 4. There is no valid array of smaller length.

Notice that in the fourth sample the array is empty.
import java.util.*;
import java.io.*;
import java.math.*;

public class Main {

	private static FastReader fastReader = new FastReader();
	public static void main(String[] args) {
		Task solver = new Task();
		solver.solve();
	}
 
	static class Task{
		
		public void solve() {
			long u = fastReader.nextLong();
			long v = fastReader.nextLong();
			if(u>v || ((u&1) != (v&1))){
				System.out.println(-1);
				return;
			}
			else if(u==v){
				if(u==0) System.out.println(u);
				else System.out.println("1\n"+u);
				return;
			}
			long z = (v-u)/2;
			if((u & z) != 0) System.out.println("3\n"+u+" "+z+" "+z);
			else System.out.println("2\n"+(u+z)+" "+z);
		}
	}
 
	static class FastReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
 
		public FastReader() {
			reader = new BufferedReader(new InputStreamReader(System.in));
			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());
		}
 
		public long nextLong() {
			return Long.parseLong(next());
		}
		
		public String readLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return null;
		}
	}
}