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: #include <bits/stdc++.h>
  int main() {
    int32_t num_tests;
    num_tests = 1;
    std::mt19937 rng;
    for (int32_t t = 0; t < num_tests; t++) {
      int32_t n;
      std::cin >> n;
      int32_t* arr = new int32_t[n];
      for (int32_t i = 0; i < n; i++) std::cin >> arr[i];
      int32_t* min_end_to_reach = new int32_t[512];
      for (int32_t i = 1; i < 512; i++) min_end_to_reach[i] = INT32_MAX;
      min_end_to_reach[0] = 0;
      for (int32_t i = 0; i < n; i++) {
        for (int32_t j = 0; j < 512; j++) {
          int32_t want_prev = arr[i] ^ j;
          if (min_end_to_reach[want_prev] < arr[i])
            min_end_to_reach[j] = std::min(min_end_to_reach[j], arr[i]);
        }
      }
      std::vector<int32_t> ans;
      for (int32_t i = 0; i < 512; i++)
        if (min_end_to_reach[i] != INT32_MAX) ans.push_back(i);
      std::cout << ans.size() << "\n";
      for (int32_t i = 0; i < ans.size(); i++) std::cout << ans[i] << " ";
      std::cout << "\n";
    }
    return 0;
  }
  