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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class TEST {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		long u = Long.parseLong(st.nextToken());
		long v = Long.parseLong(st.nextToken());
		if ((u > v) || (u >> 1 << 1 == u ^ v >> 1 << 1 == v)) {
			System.out.println(-1);
		} else if (u == 0 && v == 0) {
			System.out.println(0);
		} else if (u == v) {
			System.out.println("1\n" + u);
		} else {
			long x = (v + u) >> 1;
			long y = v - x;
			if (x + y == v && ((x ^ y) == (u)))
				System.out.printf("2\n%d %d", x, y);
			else
				System.out.printf("3\n%d %d %d", u, (v - u) >> 1, (v - u) >> 1);
		}
	}

}