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 kset(n, k): 
	if ((n >> (k - 1)) and 1): 
		return 1 
	else: 
	    return 0
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] 
ss=input().split()
x=int(ss[0])
s=int(ss[1])
if x==s==0:
    print(0)
elif s==x and x!=0 and s!=0:
    print(1)
    print(s)
elif s<x or (s-x)%2 !=0:
    print(-1)
else:
    z=compute(s,x)
    if z[0]==-1:
        print(3)
        print((s-x)//2,(s-x)//2,x)
    else:
        print(2)
        print(z[0],z[1])
    
        
        