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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class D {
    final int INF = 0x3f3f3f3f;
    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;

    public D() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) {
        new D();
    }

    private void solve() {
        long a = nextLong();
        long b = nextLong();
        if (a == b && a == 0) {
            System.out.println(0);
            return;
        }
        if(a == b) {
            System.out.println(1);
            System.out.println(b);
            return;
        }
        if (a > b) {
            System.out.println(-1);
            return;
        }
        if ((b - a) % 2 == 0) {
            long d = (b - a) / 2;
            if ((d & a) == 0) {
                outln(2);
                outln(d + " " + (a | d));
            } else {
                outln(3);
                out(d + " " + d + " ");
                outln(a);
            }
        } else {
            System.out.println(-1);
        }
    }


    private void outln(Object o) {
        out.println(o);
    }

    private void out(Object o) {
        out.print(o);
    }

    public long[] nextLongArr(int n) {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextLong();
        }
        return res;
    }

    public int[] nextIntArr(int n) {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }

    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }

    public int nextInt() {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() {
        return Double.parseDouble(nextToken());
    }


}