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.
from sys import stdin



def compute(S, X): 

    A = (S - X)//2

    a = 0

    b = 0

  

    # Traverse through all bits 

    for i in range(64): 

        Xi = (X & (1 << i)) 

        Ai = (A & (1 << i)) 

        if (Xi == 0 and Ai == 0): 

            # Let us leave bits as 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)  

            # We leave i-th bit of b as 0. 

  

        else: # (Xi == 1 and Ai == 1) 

            return -1

  

    return (a,b)




def main():
	inp=stdin.readline
	u,v=map(int, inp().split())
	if u==0 and v==0:
		print(0)
	elif u>v or (v-u)%2 !=0:
		print("-1")
	elif (v-u)==0:
		print(1)
		print(u)
	else:
		f=compute(v,u)
		if f !=-1:
			print(2)
			print(f[0],f[1])
		else:		
			a=u
			b=(v-u)//2
			print(3)
			print(a,b,b)


main()