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.
def two(s, x):
    aa = (s - x) // 2; 
    a = 0
    b = 0;  
    for i in range(0, 100):
        nowi = (x & (1 << i)) 
        nowa = (aa & (1 << i))
        if (nowi == 0 and nowa == 0):
            continue;
        if (nowi == 0 and nowa > 0):
            a = ((1 << i) | a)  
            b = ((1 << i) | b)
        elif (nowi > 0 and nowa == 0):
            a = ((1 << i) | a);  
        else:
            return (0, 0) 
    return (a, b)
def main():
	u, v = [int(i) for i in input().split()]
	if ((v - u) % 2 != 0):
	    print(-1)
	    return 0
	if (u == v):
		if (u == 0):
			print(0)
			return 0
		else:
			print(1, v)
			return 0
	if (u > v):
		print(-1)
		return 0
	if (two(v, u) != (0, 0)):
		print(2)
		print(two(v, u)[0], two(v, u)[1])
		return 0;
	if ((v - u) % 2 != 0):
		print(-1)
		return 0
	print(3)
	print((v - u) // 2, (v - u) // 2, u)
main()