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.*;

import static java.lang.Math.*;

public class Main {
    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(System.out);
        long u = nextLong();
        long v = nextLong();
        if (u == 0 && v == 0) pw.println(0);
        else if (u == 0) {
            if (v % 2 == 0) {
                pw.println(2);
                pw.println((v / 2) + " " + (v / 2));
            } else pw.println(-1);
        } else if (u > v) pw.println(-1);
        else if (u == v) {
            pw.println(1);
            pw.println(u);
        } else {
            if ((v - u) % 2 != 0) pw.println(-1);
            else {
                long c = (v - u) / 2;
                long d = u + c;
                if ((d ^ c) == u) {
                    pw.println(2);
                    pw.println(c + " " + d);
                }else{
                    pw.println(3);
                    pw.println(c + " " + u + " "+ c);
                }
            }
        }
        pw.close();
    }

    static long mod = (long) 1e9 + 7;

    static PrintWriter pw;
    static StringTokenizer st = new StringTokenizer("");
    static BufferedReader br;

    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

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

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

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