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 statistics
import math
import datetime
import collections

def identity(*args):
    if len(args) == 1:
        return args[0]
    return args


def parsin(*, l=1, vpl=1, cf=identity, s=" "):
    if l == 1:
        if vpl == 1:
            return cf(input())
        else:
            return list(map(cf, input().split(s)))
    else:
        if vpl == 1:
            return [cf(input()) for _ in range(l)]
        else:
            return [list(map(cf, input().split(s)))
                    for _ in range(l)]

def odw(l):
    wyn = 0
    pot = 1
    for i in l:
        wyn += pot * i
        pot *= 2
    return wyn


def konw(n):
    l = []
    while n > 0:
        l.append(n % 2)
        n //= 2
    return l

def main():
    x, s = map(int, input().split())
    xx = konw(x)
    ss = konw(s)
    if x == 0 and s == 0:
        print(0)
    else:
        if x > s:
            print(-1)
        if x == s:
            print(1)
            print(x)
        if x < s:
            if (x-s) % 2 == 1:
                print(-1)
            else:
                kan1 = []
                kan2 = []
                for i in range(len(xx)):
                    if xx[i] == 0:
                        kan1.append(0)
                        kan2.append(0)
                    else:
                        kan1.append(1)
                        kan2.append(0)
                for j in range(len(xx), len(ss)):
                    kan1.append(0)
                    kan2.append(0)
                ind = len(ss)-1
                while odw(kan1) + odw(kan2) < s:
                    while ind >= 0:
                        if not (kan1[ind] == kan2[ind] and kan1[ind] == 0):
                            ind -= 1
                        else:
                            if 2**(ind+1)+odw(kan1)+odw(kan2) > s:
                                ind -= 1
                            else:
                                break
                    if ind == -1:
                        break
                    kan1[ind] = 1
                    kan2[ind] = 1
                if odw(kan1)+odw(kan2) == s:
                    print(2)
                    print(odw(kan1), odw(kan2))
                else:
                    print(3)
                    print(x, (s-x)//2, (s-x)//2)

    pass


if __name__ == '__main__':
    main()


