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.*;
  
  public class HelloWorld {
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int t = 1;
          StringBuilder ret = new StringBuilder();
          while(t-- > 0) {
              int n = sc.nextInt();
              int[] a = new int[n];
              for(int i = 0; i < n; i++) a[i] = sc.nextInt();
              int[] dp = new int[514];
              Arrays.fill(dp, 600);
              dp[0] = 0;
              for(int i = 0; i < n; i++) {
                  for(int j = 0; j <= 512; j++) {
                      if(dp[j] < a[i]) {
                          dp[j ^ a[i]] = Math.min(dp[j ^ a[i]], a[i]);
                      }
                  }
              }
              ArrayList<Integer> arr = new ArrayList<>();
              for(int i = 0; i <= 512; i++) {
                  if(dp[i] < 600) arr.add(i);
              }
              ret.append(arr.size() + "\n");
              for(int aa:arr) ret.append(aa + " ");
              ret.append("\n");
          }
          System.out.println(ret);
      }
  }