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

public class Main
{
	public static void main(String[] args) {
		new Thread(null, new Runnable() {
			public void run() {
                solve();
            }
        }, "1", 1 << 26).start();
	}
	static void solve () {
		FastReader fr =new FastReader();	PrintWriter op =new PrintWriter(System.out);
 
 		long u =fr.nextLong() ,v =fr.nextLong() ,w ,i ,j ,k ,l ,a ,b ;
 		if (u>v || ((u&1) != (v&1)))	op.println("-1") ;
 		else if (u==v && u==0)	op.println("0") ;
 		else if (u==v)	op.println("1\n"+u) ;
 		else {
 			w =(v-u)/2 ;	i =Math.max(u,w) ;	j =1 ;	a =b =0 ;
 			while (j<=i) {
 				k =u&j ;	l =w&j ;	if (l!=0 && k!=0)	break;
 				if (l!=0)	{	a|=j ;	b|=j ;	}
 				else if (k!=0)	a|=j ;
 				j<<=1 ;
 			}
 			if (i<j)	op.println("2\n"+a+" "+b) ;
 			else 	op.println("3\n"+u+" "+w+" "+w) ;
 		}
		op.flush();	op.close();
	}
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br =new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st==null || (!st.hasMoreElements())) 
			{
				try
				{
					st =new StringTokenizer(br.readLine());
				}
				catch(IOException e)
				{
					e.printStackTrace();
				}
				
			}
			return st.nextToken();
		}

		String nextLine() {
			String str ="";

			try
			{
				str =br.readLine();
			}
			catch(IOException e)
			{
				e.printStackTrace();
			}

			return str;
		}

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

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