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.*;
public class GFG {
	public static void main (String[] args) 
	{
		Scanner sc=new Scanner(System.in);
		long a[]=new long[70];long b[]=new long[70];int t;long s,i,j,k;
		long u=sc.nextLong();long v=sc.nextLong();boolean b1=false;
		if (u==0 && v==0) System.out.println(0);
		else if (u>v)System.out.println(-1);
		else if (u==v) System.out.println(1+"\n"+v);
		else 
		{
		    t=0;i=u;
		    while (u>0)
		    {
		        a[t++]=u%2;
		        u/=2;
		    }
		    s=v-i;//System.out.println(s);
		    if (s%2==1){System.out.println(-1);return;}
		    s/=2;t=0;j=s;
		    //System.out.println(s);
		    while (s>0)
		    {
		        b[t++]=s%2;
		        s/=2;
		    }
		    for (int i1=0;i1<70;i1++)
		    if (b[i1]==1 && a[i1]==1)b1=true;
		    if (b1)
		    {
		        System.out.println(3);
		        System.out.println(i+" "+j+" "+j);
		    }
		    else 
		    {
		        System.out.println(2);
		        System.out.println((i+j)+" "+j);
		    }
		}
	}
}