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 pr1(a):
    print(a)
    exit()    
    
flag = [0] * 100

u, v = map(int, input().split())

if u <= v:
    if v == 0:
        pr1(0)
        
    if v == u:
        print(1)
        print(v)
        exit()
    
    if (v - u) % 2 == 1:
        pr1(-1)
        
    fl = True
    
    for i in range(70):
        if (v - u) & (1 << i):
            flag[i - 1] += 1
            if u & (1 << (i - 1)):
                fl = False
                
    result = [u, 0]
    for i in range(70):
        if flag[i]:
            result[1] += 1 << i    
    
    if fl:
        for i in range(70):
            if flag[i]:
                result[0] += 1 << i
                
    else:
        result.append(0)
        for i in range(70):
            if flag[i]:
                result[2] += 1 << i  
                
    print(len(result))
    print(*result[::-1])
    
else:
    pr1(-1) 