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 tank {
  
      static final FastScanner fs = new FastScanner();
  
      static PrintWriter out = new PrintWriter(System.out);
  
      public static void main(String[] args) {
          int t = 1;
          while(t-- > 0) run_case();
  
          out.close();
      }
  
      static void run_case() {
          int n = fs.nextInt();
          int[] a = fs.readArray(n);
          int[][] cnt = new int[513][2];
  
          cnt[0][0] = 1;
          cnt[0][1] = 0;
  
          for (int i = 0; i < n; i++) {
              for (int j = 0; j < 513; j++) {
                  if(cnt[j][0] > 0 && cnt[j][1] < a[i]) {
                      cnt[a[i] ^ j][0]++;
                      if(cnt[a[i] ^ j][0] > 1)
                          cnt[a[i] ^ j][1] = Math.min(a[i], cnt[a[i] ^ j][1]);
                      else
                          cnt[a[i] ^ j][1] = a[i];
                  }
              }
          }
  
          int k = 0;
  
          for (int i = 0; i < 513; i++) {
              if(cnt[i][0] > 0) k++;
          }
  
          out.println(k);
  
          for (int i = 0; i < 513; i++) {
              if(cnt[i][0] > 0) out.print(i + " ");
          }
  
          out.println();
      }
  
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          String nextLine(){
              try {
                  return br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
  
              return "";
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
      }
  }