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.math.BigDecimal;
import java.util.*;

public class Day9 {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(System.out);
        StringTokenizer st = new StringTokenizer(reader.readLine());
        long u = Long.parseLong(st.nextToken());
        long v = Long.parseLong(st.nextToken());
        if(u > v){
            System.out.println(-1);
            return;
        }
        else if(v == 0){
            System.out.println(0);
            return;
        }
        if(u == 0){
            if(v % 2 != 0){
                System.out.println(-1);
            }
            else{
                System.out.println(2);
                System.out.println((v / 2) + " " + (v / 2));
            }
            return;
        }
        if(u % 2 != v % 2){
            System.out.println(-1);
            return;
        }
        if(u == v){
            System.out.println(1);
            System.out.println(u);
            return;
        }
        long f = u;
        long ost = v - u;
        char[] b = Long.toBinaryString(f).toCharArray();
        long[] steps = new long[65];
        steps[0] = 1;
        for(int i = 1; i < 65; ++i){
            steps[i] = steps[i - 1] * (long)2;
        }
        boolean can = true;
        long sec = 0;
        for(int i = 61; i >= 0; --i){
            if(steps[i] * (long)2 <= ost){
                if(i >= b.length || b[b.length - 1 - i] == '0'){
                    f += steps[i];
                    sec += steps[i];
                    ost -= steps[i] * (long)2;
                }
                else{
                    can = false;
                    break;
                }
            }
        }
        if(can){
            System.out.println(2);
            System.out.println(f + " " + sec);
            return;
        }
        f = u;
        ost = v - u;
        b = Long.toBinaryString(f).toCharArray();
        System.out.println(3);
        System.out.println(f + " " + (ost / 2) + " " + (ost / 2));
        writer.close();
    }
}