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.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long u = sc.nextLong();
		long v = sc.nextLong();
		sc.close();
		if (u > v || ((v - u) & 1) != 0) {
			System.out.print(-1);
			return;
		}
		if (u == 0) {
			if (v == 0) {
				System.out.print(0);
				return;
			} else {
				System.out.printf("2\n%d %d", v / 2, v / 2);
				return;
			}
		}
		if (u == v) {
			System.out.printf("1\n%d", u);
			return;
		}
		if ((u & (v - u) / 2) != 0) {
			System.out.printf("3\n%d %d %d", u, (v - u) / 2, (v - u) / 2);
			return;
		} else {
			System.out.printf("2\n%d %d", u + (v - u) / 2, (v - u) / 2);
			return;
		}
	}
}

		 	 				   				   		 		 					