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 java.io.*;
import java.util.*;

/**
  * Built using my Brain
  * Actual solution is at the bottom
  *
  * @author Lenard Hoffstader
*/

public class cfjava
{
  public static void main(String[] args)
  {
    OutputStream outputStream = System.out;
    FastReader in = new FastReader();
    PrintWriter out = new PrintWriter(outputStream);
    PROBLEM solver = new PROBLEM();
    int t=1;
    // t=in.nextInt();
    for(int i=0;i<t;i++){
      solver.solve(in, out);
    }
    out.close();
  }

  static class PROBLEM
  {
    public void solve(FastReader in,PrintWriter out)
    {
      long u=in.nextLong();
      long v=in.nextLong();
      if(u%2!=v%2 || u>v){
        out.println(-1);
      }
      else if(u==0 && v==0){
        out.println(0);
      }
      else if(u==v){
        out.println(1);
        out.println(u);
      }
      else {
        long p=(v-u)/2;
        long q=p&u;
        if(q > 0){
          out.println(3+"\n"+u+" "+p+" "+p);
        }
        else out.println(2+"\n"+(u+p)+" "+p);
      }
    }
  }
  
  static class FastReader 
  {
    BufferedReader br; 
    StringTokenizer st; 

    public FastReader() 
    { 
      br = new BufferedReader(new InputStreamReader(System.in)); 
    } 

    String next() 
    { 
      while (st == null || !st.hasMoreElements()) 
      {
        try
        { 
          st = new StringTokenizer(br.readLine()); 
        } 
        catch (IOException  e) 
        { 
          e.printStackTrace(); 
        } 
      } 
      return st.nextToken(); 
    } 

    int nextInt(){return Integer.parseInt(next());}

    long nextLong(){return Long.parseLong(next());} 

    double nextDouble(){return Double.parseDouble(next());}

    boolean nextBoolean(){return Boolean.parseBoolean(next());}

    String nextLine() 
    { 
      String str = ""; 
      try
      { 
        str = br.readLine(); 
      } 
      catch (IOException e){e.printStackTrace();} 
      return str; 
    } 
  }
}
