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.io.*;
import java.util.*;

public class Main {
	public static void main(String[] args) {
		InputReader in = new InputReader();
		StringBuilder out = new StringBuilder();

		long u = Long.parseLong(in.next());
		long v = Long.parseLong(in.next());
		if (u > v || (v - u) % 2 == 1) {
			System.out.println(-1);
			return;
		}
		if (u == v) {
			if (u == 0) {
				System.out.println(0);
			} else {
				System.out.println(1);
				System.out.println(u);
			}
			return;
		}
		List<Long> list = new ArrayList<>();
		list.add(u);
		v -= u;
		if ((u & (v / 2)) == 0) {
			list.set(0, u | (v / 2));
		} else {
			list.add(v / 2);
		}
		list.add(v / 2);
		out.append(list.size() + "\n");
		for (long x : list) {
			out.append(x + " ");
		}
		out.setCharAt(out.length() - 1, '\n');

		System.out.print(out);
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer st;

		public InputReader() {
			reader = new BufferedReader(new InputStreamReader(System.in));
		}

		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(nextLine());
			}
			return st.nextToken();
		}

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return null;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}
