Problem: 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.

Solution (in PYTHON3):

Solution:
# Python program to find two numbers with
# given Sum and XOR such that value of
# first number is minimum.


# Function that takes in the sum and XOR
# of two numbers and generates the two
# numbers such that the value of X is
# minimized
def compute(S, X):
    A = (S - X) // 2
    a = 0
    b = 0
    if (S-X)%2!=0:
        return -1

    # 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
x,s=list(map(int,input().split()))
ans=compute(s,x)
if ans!=-1:
    if ans[0]==0 and ans[1]==0:
        print(0)
    elif ans[1]==0:
        print(1)
        print(ans[0])
    else:
        print(2)
        print(ans[0],ans[1])
else:
    
    if x==0 or x>s:
        print(-1)
    elif s%x==0 and (s//x)%2!=0:
        print(s//x)
        print(*[x]*(s//x))
    elif (s-x)%2==0 :
        print(3)
        print(x,(s-x)//2,(s-x)//2)

    else:
        print(-1)

# This code is contributed by ankush_953