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 HelloWorld{

    private static void fill(long u, int arr[]){
        int index = 0;
        while(u > 0){
            arr[index++] = (int)(u%2);
            u /= 2;
        }
    }
    
    private static int search(long v){
        int i=0;
        while(((long)2<<i)<=v) i++;
        return i-1;
    }

    public static void main(String []args)throws IOException{
    
        Scanner sc = new Scanner(System.in);
        
        long u = sc.nextLong(), v = sc.nextLong();
        int arr[] = new int[100];
    
        v -= u;
        
        if(v < 0 || v%2 == 1)   
        {
            System.out.println(-1);
            System.exit(0);
        }
        
        fill(u, arr);
        //System.out.println(Arrays.toString(arr));
        
        while(v > 0){
            int index = search(v);
            arr[index] +=2; 
            v -= ((long)2<<index);
        }
        //System.out.println(Arrays.toString(arr));
        
        int max = 0;
        for(int i=0; i<100; i++) max = Math.max(arr[i], max);
        
        long res[] = new long[max];
        
        for(int i=0; i<100; i++){
            for(int j=0; j<arr[i]; j++){
                res[j] += ((long)1<<i);
            }
        }
        
        System.out.println(+max);
        for(long z : res)
            System.out.print(+z+" ");
    }
}