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 EhabtheXorcist {
	@SuppressWarnings("resource")
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		long u = input.nextLong();
		long v = input.nextLong();
		if (u == 0 && u == v) {
			System.out.println(0);
			return;
		}
		if (u == v) {
			System.out.println("1\n" + u);
			return;
		}
		if (u % 2 != v % 2 || u > v) {
			System.out.println(-1);
			return;
		}
		long and = (v - u) / 2;
		long c = u + and, d = and;
		if ((c ^ d) == u) {
			System.out.println("2\n" + c + " " + d);
			return;
		}
		System.out.println("3\n" + (v - u) / 2 + " " + (v - u) / 2 + " " + u);
		input.close();
	}
}