description: 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] 
solution: import java.io.*;
  import java.util.*;
  
  public class Codeforces
  {
      public static void main(String args[])throws Exception
      {
          BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
          StringBuilder sb=new StringBuilder();
          int n=Integer.parseInt(bu.readLine());
          String s[]=bu.readLine().split(" ");
          int a[]=new int[n],i;
          for(i=0;i<n;i++) a[i]=Integer.parseInt(s[i]);
  
          int N=501,M=512,j;
          int min[]=new int[M];
          Arrays.fill(min,N);
  
          min[0]=0;
          for(i=0;i<n;i++)
          for(j=0;j<M;j++)
          //check if a smaller number could be used to xor to this value
          if(min[j]<a[i]) min[j^a[i]]=Math.min(min[j^a[i]],a[i]);
  
          ArrayList<Integer> ans=new ArrayList<>();
          for(i=0;i<M;i++)
          if(min[i]<N)ans.add(i);
          sb.append(ans.size()+"\n");
          for(int x:ans) sb.append(x+" ");
          System.out.print(sb);
      }
  }