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 = list(map(int, input().split()))
ind = 0
if u > v or (v - u) % 2 == 1:
        print(-1)
elif v == 0:
        print(0)
elif u == v:
        print("1")
        print(u)
else:
        v = (v - u) // 2
        bit_u = [0] * 100
        bit_v = [0] * 100
        ind = 0
        while v > 0:
                if v % 2 == 1:
                        bit_v[ind] = 1;
                v = v // 2
                ind += 1
        ind = 0
        while u > 0:
                if u % 2 == 1:
                        bit_u[ind] = 1;
                u = u // 2
                ind +=  1
        b = 1
        for i in range(0, 100):
                if bit_u[i] + bit_v[i] == 2:
                        b = 0
        if b == 1:
                print(2)
                a1 = 0
                a2 = 0
                for i in range(0, 100):
                        if bit_v[i] == 1:
                                a1 += (1 << i)
                                a2 += (1 << i)
                        if bit_u[i] == 1:
                                a2 += (1 << i)
                print(str(a1) + " " + str(a2))
        else:
                print(3)
                a1 = 0 
                a2 = 0
                a3 = 0
                for i in range(0, 100):
                        if bit_u[i] + bit_v[i] == 2:
                                a1 += (1 << i)
                                a2 += (1 << i)
                                a3 += (1 << i)
                        elif bit_v[i] == 1:
                                a1 += (1 << i)
                                a2 += (1 << i)
                        elif bit_u[i] == 1:
                                a2 += (1 << i)
                print(str(a1) + " " + str(a2) + " " + str(a3));