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.*;
// tau is superior

public class D
{
    void solve(FastIO io)
    {
        long a = io.nextLong();
        long b = io.nextLong();

        if (b < a || (a & 1) != (b & 1))
        {
            io.println(-1);
            return;
        }

        // if (a == b)
        // {
        //     io.println(1);
        //     io.println(a);
        //     return;
        // }

        

        long get = b - a;
        long first = get / 2;
        long sec = get - first;
        long curr = first ^ sec;
        long need = a ^ curr;

      
        
        ArrayList<Long> ans = new ArrayList<>();
        if (first != 0)
            ans.add(first);
        if (sec != 0)
            ans.add(sec);
        if (need != 0)
            ans.add(need);

        if ((get & 1) == 0 && ((get >> 1) & a) == 0)
        {
            ArrayList<Long> small = new ArrayList<>();

            first = a | (get >> 1);
            sec = get >> 1;

            if (first != 0)
                small.add(first);
            if (sec != 0)
                small.add(sec);

            if (small.size() < ans.size())
                ans = small;
        }

        io.println(ans.size());
        for (long i : ans)
            io.print(i + " ");
        io.println();
    }

    public static void main(String[] args) {
        FastIO io = new FastIO();

        new D().solve(io);

        io.close();
    }

    static class FastIO extends PrintWriter
    {
        StringTokenizer st = new StringTokenizer("");
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));

        FastIO()
        {
            super(System.out);
        }

        String next()
        {
            while (!st.hasMoreTokens())
            {
                try {
                    st = new StringTokenizer(r.readLine());
                } catch (Exception e) { }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }
    }
}