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.Scanner;

public class train {

    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        long u = in.nextLong();
        long v = in.nextLong();
        boolean b = false;
        long x;
        if (u == 0 && v == 0) {
            System.out.println("0");
            b = true;
        } else if (u > v) {
            b = false;
        }
        else if(u == v){
            System.out.println("1\n"+u);b= true;
        }else {
            if (u == 1) {
                b = true;
                System.out.println(v);
                for (int i = 0; i < v; i++) {
                    System.out.print("1 ");
                }
            } else if ((v - u) % 2 == 0) {
                x = (v - u) / 2;
                if (x + (x ^ u) == v) {
                    System.out.println("2\n" + x + " " + ((x ^ u)));
                } else {
                    System.out.println("3\n" + x + " " + x + " " + u);
                }
                b = true;
            }

        }
        if (b == false) {
            System.out.println("-1");
        }

    }

}