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 f(u,v):
    x=[int(y) for y in bin(u)[2:]]
    s=[int(z) for z in bin(v)[2:]]
    x=[0]*(len(s)-len(x))+x
    a=[0]*len(x)
    b=[0]*len(x)
    for i in range(len(x) - 1, 0, -1):
        if x[i] == 1:
            a[i] = 0
            b[i] = 1
        else:
            if s[i - 1] == x[i - 1]:
                a[i] = 0
                b[i] = 0
            else:
                a[i] = 1
                b[i] = 1
    if x[0] == 1:
        a[0] = 0
        b[0] = 1
    c=0
    d=0
    for i in range(len(a)):
        c+=pow(2,len(a)-i-1)*a[i]
        d+=pow(2,len(a)-i-1)*b[i]
    return c,d

u,v=map(int,input().split())
if u % 2 != v % 2 or v < u:
    print(-1)
elif u==0 and v==0:
    print(0)
elif u==v:
    print(1)
    print(u)
else:
    c,d=f(u,v)
    if c+d==v and c^d==u:
        print(2)
        print(c,d)
    else:
        print(3)
        print(u, (v - u)//2, (v - u)//2)
