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.*;

public class xorcist {

    static int[] setctr;

    public static void main(String[] args) throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        StringTokenizer st = new StringTokenizer(f.readLine());

        long u = Long.parseLong(st.nextToken());
        long v = Long.parseLong(st.nextToken());

        if(u > v){pw.println(-1); pw.close(); return;}
        if(u==0 && v == 0){pw.println(0); pw.close(); return;}
        if(u == v){pw.println(1); pw.println(u); pw.close(); return;}


        long vmu = v-u;
        String ustr = Long.toBinaryString(u);
        String vstr = Long.toBinaryString(v);
        String vmustr = Long.toBinaryString(vmu);

        int ulen = ustr.length();
        int vlen = vstr.length();
        int vmulen = vmustr.length();
        setctr = new int[vlen];


        for(int i = 1; i <= ulen; i++){
            setctr[vlen-i]+=(ustr.charAt(ulen-i)-'0');
        }
        for(int i = 1; i <= vmulen; i++){
            if(vmustr.charAt(vmulen-i)=='1'){
                if(i==1){
                    pw.println(-1); pw.close(); return;
                } else {
                    setctr[vlen-i+1]+=2;
                }
            }
        }

        int maxval = 0;
        for(int i = 0; i < setctr.length; i++){
            maxval = Math.max(maxval, setctr[i]);
        }
        pw.println(maxval);
        while(maxval != 0){
            String str = "";
            maxval = 0;
            for(int i = 0; i < setctr.length; i++){
                if(setctr[i]>=1){
                    str+="1"; setctr[i]--;
                } else {str+="0";}
                maxval=Math.max(maxval, setctr[i]);
            }
            pw.print(Long.parseLong(str, 2));
            if(maxval!=0) pw.print(" ");
        }
        pw.println();




        pw.close();
    }

}
