This is an easier version of the problem with smaller constraints.

Korney Korneevich dag up an array a of length n. Korney Korneevich has recently read about the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR), so he wished to experiment with it. For this purpose, he decided to find all integers x ≥ 0 such that there exists an increasing subsequence of the array a, in which the bitwise XOR of numbers is equal to x.

It didn't take a long time for Korney Korneevich to find all such x, and he wants to check his result. That's why he asked you to solve this problem!

A sequence s is a subsequence of a sequence b if s can be obtained from b by deletion of several (possibly, zero or all) elements.

A sequence s_1, s_2, … , s_m is called increasing if s_1 < s_2 < … < s_m.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 500) — the elements of the array a.

Output

In the first line print a single integer k — the number of found x values.

In the second line print k integers in increasing order x_1, x_2, … x_k (0 ≤ x_1 < … < x_k) — found x values.

Examples

Input


4
4 2 2 4


Output


4
0 2 4 6 


Input


8
1 0 1 7 12 5 3 2


Output


12
0 1 2 3 4 5 6 7 10 11 12 13 

Note

In the first test case:

  * To get value x = 0 it is possible to choose and empty subsequence 
  * To get value x = 2 it is possible to choose a subsequence [2] 
  * To get value x = 4 it is possible to choose a subsequence [4] 
  * To get value x = 6 it is possible to choose a subsequence [2, 4] 
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.StringTokenizer;


import java.io.*;
 
public class cf745 {
	
	private static class MyScanner {
	    BufferedReader br;
	    StringTokenizer st;
 
	    public MyScanner() {
	       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());
	    }
 
	    String nextLine(){
	        String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
	    }
 
	 }

	public static void solution(int n, int[] arr)	
	{
		int[] ans = new int[512];
		for(int i = 0 ;i<512; i++)
			ans[i] = -1;
		
		for(int i = n-1; i>=0; i--)
		{
			int curr = arr[i];
			ans[curr] = Math.max(ans[curr], curr);
			
			for(int j = 0; j<512; j++)
			{
			   if(ans[j]>curr)
				   ans[j^curr] = Math.max(ans[j^curr], curr);
			}
		}
		
		ans[0] = 1;
		int count = 0;
		
		for(int i = 0; i<512; i++)
			{
			    if(ans[i]>=0)
			    	count++;
			}
		
		out.println(count);
		
		for(int i = 0; i<512; i++)
			{
			    if(ans[i]>=0)
			      out.print(i+" ");
			}
		
	}
	
	
private static PrintWriter out = new PrintWriter(System.out);
 
public static void main (String[] args)
{
	MyScanner s =  new MyScanner();
    
	int n = s.nextInt();
		int[] arr = new int[n];
		
		for(int i = 0; i<n; i++)
			arr[i] = s.nextInt();
 		
        solution(n,arr);
    
     
    out.flush();
    out.close();
}
}