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.util.Scanner;
  
  public class F {
      public static void main(String[] args) {
          Scanner s = new Scanner(System.in);
  
          solve(s.nextInt(), s);
      }
  
  
      public static void solve(int n, Scanner s) {
          int min[] = new int[512];
          for (int i = 0; i < 512; i++) {
              min[i] = -1;
          }
          min[0] = -2;
  
          int t;
          for (int i = 0; i < n; i++) {
              t = s.nextInt();
  
              for (int j = 0; j < 512; j++) {
                  if (min[j] == -1) continue;
                  if (min[j] > t) continue;
                  if (min[j ^ t] != -1) min[j ^ t] = Math.min(t, min[j ^ t]); else min[j ^ t] = t;
              }
          }
  
          String erg = "";
          int sol = 0;
  
          for (int i = 0; i < 512; i++) {
              if (min[i] != -1) {
                  sol++;
                  erg += i + " ";
              }
          }
  
          System.out.println(sol);
          System.out.println(erg);
      }
  }
  