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.
#!usr/bin
import math
if __name__=="__main__":
    u, v=map(int, input().split()) 
    if v!=0: #
        k=math.ceil(math.log2(v))
    if u==0 and v==0:
        print(0)
        exit(0)
    if u==v: #
        print(1)
        print(v)
        exit(0)
    if u>v or (u^v)&1==1:
        print(-1)
    else:
        carry=0
        able=True
        a=b=0
        for i in range(k+1):#+1wei
            xor=(u>>i)&1
            sum=(v>>i)&1
            if carry==0:
                if sum^xor==1:
                    able=False
                    break
                elif sum==1:
                    a=a|(1<<i)
                    b=b&~(1<<i)
                else:
                    carry=1
            elif carry==2:
                if sum^xor==0:
                    able=False
                    break
                elif xor==1:
                    a=a|(1<<i)
                    b=b&~(1<<i)
                    carry=2
                else:
                    carry=1
            else:
                if xor^sum==0:
                    a=a&~(1<<(i-1))
                    b=b&~(1<<(i-1))
                else:
                    a=a|(1<<(i-1))
                    b=b|(1<<(i-1))
                if xor==1:
                    a=a|(1<<i)
                    b=b&~(1<<i)
                    if sum==0: #
                        carry=2
                    else:
                        carry=0
                else:
                    carry=1
                
        if able==True:
            print(2)
            print("%d %d"%(a,b))
        else:
            v = v-u
            for i in range(k):
                if (v>>(i+1))&1==1:
                    v=v|(1<<i)
                elif (v>>i)&1==1:
                    v=v^(1<<i)
            print(3)
            print(" ".join((str(i) for i in (u,v,v))))
