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 q5 {
      public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  
      public static void solve() throws Exception {
          String[] parts = br.readLine().split(" ");
          int n = Integer.parseInt(parts[0]);
  
          Integer[] arr = new Integer[n];
  
          parts = br.readLine().split(" ");
          for (int i = 0; i < n; i++) {
              arr[i] = Integer.parseInt(parts[i]);
          }
  
          int[] vis = new int[513];
          Arrays.fill(vis,Integer.MAX_VALUE);
          vis[0] = 0;
  
          for(int i = 0;i < n;i++){
              for(int j = 0;j < vis.length;j++){
                  if(vis[j] < arr[i]){
                      vis[arr[i] ^ j] = Math.min(vis[arr[i] ^ j],arr[i]);
                  }
              }
          }
  
          int count = 0;
          for(int val : vis) if(val != Integer.MAX_VALUE) count++;
  
          System.out.println(count);
          for(int i = 0;i < vis.length;i++){
              if(vis[i] != Integer.MAX_VALUE) System.out.print(i + " ");
          }
          System.out.println();
      }
  
      public static void main(String[] args) throws Exception {
  
          // int tests = Integer.parseInt(br.readLine());
          // for (int test = 1; test <= tests; test++) {
              solve();
          // }
      }
  }
  