Problem: 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.

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {
    private static final Scanner sc = new Scanner(System.in);
    private static int[] readIntArray(Scanner s, int n) {
        int[] ans = new int[n];
        for (int i = 0; i < n; i++) {
            ans[i] = sc.nextInt();
        }
        return ans;
    }
    static int maxPower;
    static long[] power;
    static {
        long x = (long) Math.pow(10, 18);
        maxPower = 0;
        while (x > 1) {
            maxPower++;
            x /= 2;
        }
        power = new long[maxPower + 1];
        power[0] = 1;
        for (int i = 1; i <= maxPower; i++)
            power[i] = 2 * power[i-1];
    }

    static void addPowers(int[] count, long u, int factor) {
        int t = 0;
        while (u > 0) {
            if (u % 2 == 1) count[t] += factor;
            u /= 2;
            t++;
        }
    }
    static int max(int[] a) {
        int ans = 0;
        for (int x : a) ans = Math.max(ans, x);
        return ans;
    }


    
    public static void main(String[] args) {
        long u = sc.nextLong();
        long v = sc.nextLong();
        int[] count = new int[maxPower + 1];

        v = v - u;
        if (v < 0 || v % 2 == 1) System.out.println(-1);
        else {
            addPowers(count, u, 1);
            addPowers(count, v/2, 2);
            int n = max(count);
            long[] ans = new long[n];
            for (int i = 0; i < count.length; i++) {
                for (int j = 0; j < count[i]; j++) {
                    ans[j] += power[i];
                }
            }

            System.out.println(n);
            for (long x : ans) System.out.println(x + " ");

            System.out.println();
            
        }

        
    }
}