Problem: 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.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class Solution{
	
	static long mod = 998244353L;
	
	public static void main(String[] args) throws Exception{

		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		
		long u = fs.nextLong();
		long v = fs.nextLong();
		
		if(u==0 && v==0) {
			out.println(0);
			out.close();
			return;
		}

		long n = v - u;
		
		if(n<0 || (n&1)==1) {
			out.println(-1);
			out.close();
		}
		
		n = n/2;
		
		if(n==0) {
			out.println("1");
			out.println(u);
		}
		else if(((~u)&n)==n) {
			out.println(2);
			out.println((u|n)+" "+n);
		}
		else {
			out.println(3);
			out.println(u+" "+n+" "+n);
		}
		
		out.close();
	}
	
	
	
	
	static class FastScanner{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		public String next(){
			while(!st.hasMoreElements()){
				try{
					st = new StringTokenizer(br.readLine());
				} catch(IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		public String nextLine() throws IOException {
			return br.readLine();
		}
		
		public int nextInt(){
			return Integer.parseInt(next());
		}

		public int[] readArray(int n){
			int[] a = new int[n];
			for(int i=0;i<n;i++)
				a[i] = nextInt();
			return a;
		}
		
		public long nextLong() {
			return Long.parseLong(next());
		}
	}
	
}