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.ArrayList;
import java.util.Arrays;

public class Ehab_the_Xorcist {

    public static void main(String []args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String []str = br.readLine().split(" ");
        long u = Long.parseLong(str[0]);
        long v = Long.parseLong(str[1]);
        if(((v-u)%2!=0) || v<u){
            System.out.println("-1");
            return;
        }
        if(u==v && u!=0){
            System.out.println(1);
            System.out.println(v);
            return;
        }
        if(u==0 && v==0){
            System.out.println(0);
            return;
        }
        long x = (v - u)/2;
        if((x&u)!=0L){
            System.out.println(3);
            System.out.println(u+" "+x+" "+x);
        }else{
            System.out.println(2);
            System.out.println((u+x)+" "+x);
        }
    }

}