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

if (u - v) % 2 != 0 or u > v:
    print(-1)
elif u == 0 == v:
    print(0)
elif u == v:
    print(1)
    print(u)
else:
    U = u
    V = v
    u = bin(u)[2:]
    v = bin(v)[2:]

    u = "0" * (70 - len(u)) + u
    v = "0" * (70 - len(v)) + v

    A = [0] * 70
    b = True
    for i in range(70):
        # print(i)
        if u[i] == '0' and v[i] == '0':
            continue
        if u[i] == '0' and v[i] == '1':
            if i == 69:
                print(-1)
                exit(0)
            A[i+1] += 2
        if u[i] == '1' and v[i] == '1':
            A[i] += 1
        if u[i] == '1' and v[i] == '0':
            if i == 69:
                b = False

            flag = False
            for j in range(i, -1, -1):
                if A[j] == 0:
                    A[j] = 2
                elif A[j] == 1:
                    break
                else:
                    A[j] -= 2
                    flag = True
                    break
            if not flag:
                b = False
            A[i] += 1
            A[i+1] = 2
            continue

    # print(A)
    if max(A) >= 3 or not b:
        print(3)
        print((V-U)//2, (V-U)//2, U)
        exit(0)
    print(max(A))
    for i in range(1,1+max(A)):
        num = 0
        for j in range(1,70):
            if A[-j] >= i:
                num += 2**(j-1)
        print(num, end=" ")