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 PYTHON):

Solution:
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math

# sys.stdin = open('input')

def inp():
    return map(int, raw_input().split())

def main():
        u, v = inp()
        if u>v:
            print -1
            return
        if u==v:
            if u==0:
                print 0
            else:
                print 1
                print u
            return
        bu, bv = list(bin(u)[2:]), list(bin(v)[2:])
        bu = ['0']*(len(bv)-len(bu)) + bu
        # print bu, bv
        diff = v-u
        if diff%2 == 1:
            print -1
            return
        bdiff = list(bin(diff)[2:])
        bdiff = ['0']*(len(bv)-len(bdiff)) + bdiff
        bans = ['0'] * len(bv)
        for i in range(len(bv)):
            if bdiff[i] == '1':
                if i!=len(bv)-1 and bu[i+1]=='0':
                    bans[i+1] = bu[i+1] ='1'
                else:
                    break
        else:
            print 2
            print int(''.join(bu), 2), int(''.join(bans), 2)
            return
        print 3
        print u, diff/2, diff/2




main()