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 DLevelQuesXORandSUM {

	public static void main(String[] args) {
		Scanner scn=new Scanner(System.in);
		long u=scn.nextLong();
		long v=scn.nextLong();
		if(u==0)
		{
			if(v==0)
				System.out.println("0");
			else if(v%2==0)
			{
				System.out.println("2");
				System.out.println(v/2+" "+v/2);
			}else {
				System.out.println("-1");
			}
		}
		else if(u>v || u%2!=v%2)
			System.out.println("-1");
		else if(u==v) {
			System.out.println("1");
			System.out.println(u);
		}
		else {
			long x=(v-u)/2;
			if((x&u)!=0) {
				System.out.println("3");
				System.out.println(u+" "+x+" "+x);
			}else {
				System.out.println("2");
				System.out.println((u+x)+" "+x);
			}
			}
		}

}