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.util.*;
import java.io.*;

public class P1325D {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] input = br.readLine().split(" ");
        long u = Long.parseLong(input[0]);
        long v = Long.parseLong(input[1]);
        if((u%2) != (v%2) || (u > v)) {
            System.out.println(-1);
        } else if(u == v) {
            if(u == 0) {
                System.out.println(0);
            } else {
                System.out.println(1);
                System.out.println(u);
            }
        } else {
            long rem = v;
            long a = 0;
            long b = 0;
            for (int i = 60; i >= 0; i--) {
                long shift = ((long) 1) << i;
                if ((shift & u) > 0) {
                    rem -= shift;
                    a += shift;
                } else if (rem >= (shift << 1)) {
                    rem -= (shift << 1);
                    a += shift;
                    b += shift;
                }
            }
            if (rem == 0) {
                System.out.println(2);
                System.out.println(a + " " + b);
            } else {
                long o = (v - u) >> 1;
                System.out.println(3);
                System.out.println(u + " " + o + " " + o);
            }
        }
    }
}
