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.io.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
 
public class Main
{
	public static void main(String[] args) throws IOException
	{
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
		long num1 = in.nextLong();
		long num2 = in.nextLong();
		if(num1 > num2)
		{
			out.println(-1);
		}
		else 
		{
			long num3 = num2 - num1;
			if(num3 % 2 == 1)
			{
				out.println(-1);
			}
			else 
			{
				if(num1 == num2)
				{
					if(num1 == 0)
					{
						out.println(0);
					}
					else 
					{
						out.println(1);
						out.println(num2);
					}
				}
				else 
				{
				    if(((num2 - (num3 / 2)) ^ (num3 / 2)) == num1)
				    {
				        out.println(2);
				        out.println((num2 - (num3 / 2)) + " " + (num3 / 2));
				    }
				    else 
				    {
				        out.println(3);
					    out.println((num3 / 2) + " " + (num3 / 2) + " " + num1);
				    }
					
				}
			}
		}
		out.flush();
	}
}