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.util.concurrent.TimeUnit;

public class Xorcist {
	
	static long setBitNumber(long n) 
	{ 
	    if (n == 0) 
	        return 0; 
	  
	    long msb = 0; 
	    while (n != 0) 
	    { 
	        n = n / 2; 
	        msb++; 
	    } 
	  
	    return msb; 
	}
	
	static boolean isKthBitSet(long n, long k) 
	{ 
		if (((n >> (k - 1)) & 1) == 1) 
			return true;
		return false;
	}
	
	public static void main (String [] args) throws Exception {
	    BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	    StringTokenizer st = new StringTokenizer(f.readLine());
	    long u = Long.parseLong(st.nextToken());
	    long v = Long.parseLong(st.nextToken());
	    
	    if(v < u || (v-u) % 2 == 1) {
	    	out.println(-1);
	    } else if(v == u && u != 0) {
	    	out.println(1);
	    	out.println(u);
	    } else {
	    	long difference = v-u;
	    	
	    	long sbn = setBitNumber(difference/2);
	    	boolean works = true;
	    	for(long t = 1; t <= sbn; t++) {
	    		if(isKthBitSet(difference/2, t) && isKthBitSet(u, t)) {
	    			works = false;
	    		}
	    	}
	    	
	    	if(works) {
	    		if(v == 0)
	    			out.println(0);
	    		else {
	    			out.println(2);
	    			out.println((u + difference/2) + " " + (difference/2));
	    		}
	    	} else {
	    		out.println(3);
	    		out.println(u + " " + (difference/2) + " " + (difference/2));
	    	}
	    }
	    
	    out.close();
	    f.close();
	}
	
}