Problem: 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.

Solution (in JAVA):

Solution:
import java.util.*;

public class CodeForces1220D{
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		long xor = input.nextLong();
        long sum = input.nextLong();
        if(sum==xor)
        {
            if(sum!=0)
            {
                System.out.println("1");
                System.out.println(sum);
            }
            else
            System.out.println(sum);
        }
        else if(xor>sum || (sum-xor)%2==1)
        System.out.println("-1");
        else
        {
            long k=(sum-xor)/2;
            if((xor&k)==0)
            {
                System.out.println("2");
                System.out.println(k+" "+(xor+k));
            }
            else
            {
                System.out.println("3");
                System.out.println(xor+" "+k+" "+k);
            }
        }
	}
}