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.Scanner;

public class D {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        long xor = in.nextLong();
        long sum = in.nextLong();

        if (xor == sum) {
            if (xor == 0) {
                System.out.println(0);
            } else {
                System.out.println("1 \n" + xor);
            }
        } else if ((xor & 1) != (sum & 1) || sum < xor) {
            System.out.println(-1);
        } else {
            Long a = 0L, b = 0L, aC = null, bC = null;
            try {
                for (long scan = 1; scan != 0; scan <<= 1) {
                    boolean xorSet = (xor & scan) > 0;
                    boolean sumSet = (sum & scan) > 0;
                    if (sumSet) {
                        if (xorSet) {
                            a = a | scan;
                            aC = bC = null;
                        } else {
                            a = aC;
                            b = bC;
                            aC = aC | scan;
                            bC = bC | scan;
                        }
                    } else {
                        if (xorSet) {
                            a = b = null;
                            aC = aC | scan;
                        } else {
                            aC = a | scan;
                            bC = b | scan;
                        }
                    }

                }
                assert a != null;
                assert b != null;

                System.out.println("2 " + a + " " + b);
            } catch (Throwable e) {
                System.out.println("3 \n" + xor + " " + (sum - xor) / 2 + " " + (sum - xor) / 2);
            }
        }
    }
}