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.
import sys


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,-1]
    return[a,b]  

Xor,Sum = map(int,sys.stdin.readline().split())

if Xor>Sum or Xor%2!=Sum%2:
	print(-1)
elif (Xor==Sum):
	if (Sum!=0):
		print(1)
	print(Xor);
else:
	p=compute(Sum,Xor);
	if (p[0]>0 and p[1]>0):
		print(2)
		print(p[0],p[1])
	else:
		print(3)
		print(Xor,(Sum-Xor)//2,(Sum-Xor)//2)


