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.math.BigInteger;
import java.util.*;
import java.util.Map.Entry;
import java.util.Stack;


public class minjumps{
	public static class FastReader {
		BufferedReader br;
		StringTokenizer root;
		
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (root == null || !root.hasMoreTokens()) {
				try {
					root = new StringTokenizer(br.readLine());
				} catch (Exception addd) {
					addd.printStackTrace();
				}
			}
			return root.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception addd) {
				addd.printStackTrace();
			}
			return str;
		}
	}
	
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	
	public static FastReader sc = new FastReader();
 
	static int mod = (int) (1e9+7);
	static List<Integer>[] edges ;
	public static void main(String[] args) {
		
		long u = sc.nextLong() , v = sc.nextLong() ;
		
		if(u > v){
			out.print(-1);
			
		}else if(u == v){
			if(u != 0)
			out.println(1);
			out.print(v);
		}else {
		v -= u ;
		if((v&1) != 0){
			out.print(-1);
		}else {
		v /= 2 ;
		if((v^u) == (v+u)) { out.println(2); out.print(v+" "+(v^u)); }
		else { out.println(3); out.print(v+" "+v+" "+u); }
		}
		}
	out.close();
}
	
}

