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


def func(x, s):
    if s < x or abs((s-x) % 2) == 1:
#        print(-1)
        return [-1]
    elif s == x:
#        print(1)
#        print(s)
        return [s]
    else:
        andd = (s - x) // 2
#        print(andd)
        a = 0
        b = 0
        c = 0
        for i in range(64):
            xi = (x & (1 << i))
            ai = (andd & (1 << i))
            if (xi == 0 and ai == 0):
                pass
            elif (xi == 0 and ai > 0):
                a = ((1 << i) | a)
                b = ((1 << i) | b)
            elif (xi > 0 and ai == 0):
                a = ((1 << i) | a)
            else:
                c = ((1 << i) | c)
#        print([a, b, c])
        if c > 0:
            x = x ^ c
            s = s - c
            d = []
            d.extend(func(x, s))
            d.append(c)
            return d
        else:
            
            return [a, b]

a = func(x, s)
#print(len(a))
#print(a)
ok = 0
for i in a:
    if i == -1:
        print(-1)
        ok = 1
if ok == 0:
    if x == 0 and s == 0:
        print(0)
    else:
        print(len(a))    
        for i in a:
            print(i, end = ' ')



