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 EhabXorcist {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner sc = new Scanner(System.in);

		long u = sc.nextLong();
		long v = sc.nextLong();
		if (u > v) {
			System.out.println(-1);
			
		} else if (u == v) {
			if (v == 0) {
				System.out.println(0);
			} else {
				System.out.println(1);
				System.out.println(u);
			}
		} else if (u>0 && v<0 || u<0 && v>0) {
			System.out.println(-1);
		} else {

			if ((v - u) % 2 != 0) {
				System.out.println(-1);
			} else {
				long x = (v - u) / 2;

				long ans = x & u;
				if (ans == 0) {
					System.out.println(2);
					long val = u + x;
					System.out.println(val + " " + x);
				} else {
					System.out.println(3);
					System.out.println(u + " " + x + " " + x);
				}
			}
		}

	}

}