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 D {
 
	public static void main(String[] args) {
		Scanner scn = new Scanner(System.in);
		long X = scn.nextLong();
		long S = scn.nextLong();
		if (S == X) {
			System.out.println(((S == 0) ? "0" : "1" + "\n" + S));
			return;
		} else if (X > S) {
			System.out.println(-1);
			return;
		} else if (S % 2 != X % 2) {
			System.out.println(-1);
			return;
		}
		
		boolean twos = true;
		long A = (S - X) / 2;
		long a = 0, b = 0;
		final int LONG_FIELD_SIZE = 8;
 
		// Traverse through all bits
		for (int i = 0; i < (8 * LONG_FIELD_SIZE)-1; i++) {
			long sml=(long)Math.pow(2, i);
			long Xi = (X & ((long)Math.pow(2, i)));
			long Ai = (A & ((long)Math.pow(2, i)));
			if (Xi == 0 && Ai == 0) {
				// Let us leave bits as 0.
			} else if (Xi == 0 && Ai > 0) {
				a = ((sml) | a);
				b = ((sml) | b);
			} else if (Xi > 0 && Ai == 0) {
				a = ((sml) | a);
				// We leave i-th bit of b as 0.
			} else {
				twos = false;
				break;
			}
		}
		if (twos) {
			System.out.println(2);
			System.out.println(a + " " + b);
			return;
		} else {
			if (S % 2 == X % 2) {
				System.out.println(3);
				System.out.println(X + " " + (S - X) / 2 + " " + (S - X) / 2);
				return;
			}
 
		}
	}
}