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.
powers = [2**i for i in range(61)]

x, s = map(int, input().split())

if x > s:
    print(-1)
elif x == s:
    if x == 0:
        print(0)
    else:
        print(1)
        print(x)
else:
    bckt = [0]*3
    bckt[0] = x
    s -= x
    
    i = 0
    bad = set()
    good = set()
    while i < 61:
        query = x>>i
        if query%2:
            bad.add(i)
        else:
            good.add(i)
        i += 1
        
    while s > 1:
        to_add = 1
        for i in range(61):
            if powers[i] <= s//2:
                to_add = i

        s -= 2*powers[to_add]
        
        if to_add in bad:
            bckt[1] += powers[to_add]
            bckt[2] += powers[to_add]
        elif to_add in good:
            bckt[0] += powers[to_add]
            bckt[1] += powers[to_add]
                
    if s%2:
        print(-1)
    else:
        if 0 in bckt:
            to_print = [i for i in bckt if i > 0]
            print(len(to_print))
            print(*to_print)
        else:
            print(3)
            print(*bckt)