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.
import math
x,s=map(int,input().split())



if x>s:
    print(-1)
    exit

else:
    a=(s-x)//2
    p=0
    q=0
    r=0
    
    for i in range (60):
        if (x & pow(2,i)!=0) and (a & pow(2,i)==0) :
            p+=pow(2,i)
        if (x & pow(2,i)==0) and (a & pow(2,i)!=0) :
            p+=pow(2,i)
            q+=pow(2,i)
        if (x & pow(2,i)!=0) and (a & pow(2,i)!=0) :
            p+=pow(2,i)
            q+=pow(2,i)
            r+=pow(2,i)
    
    v=[]
    
    if p>0: v.append(p)
    if q>0: v.append(q)
    if r>0: v.append(r)
        
    flag=0
    if (s-x)%2==1:
        z=s-(p+q+r)
        if z^p^q^r != x :
            
            flag = 1
        else:
            v.append(z)
    
    
    if(flag == 1):
        print(-1)
    else:    
        print(len(v))
        for i in v:
            print(i,end=' ')
    
    
    
            
            