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>
  const int mod = 998244353;
  const int modd = 65536;
  const int base = 233;
  const double pi = acos(-1);
  const int N = 2e5 + 10;
  const int M = 1e9;
  const double esp = 1e-6;
  using namespace std;
  int id = 0, n, m, tot = 0, ttt = 0, a[1000010];
  vector<int> p[100010], ans;
  bool vis[100010], pis[8192][8192];
  void init() {
    for (int i = 0; i <= 8192; i++) p[i].push_back(0);
    memset(vis, 0, sizeof(vis));
    memset(pis, 0, sizeof(pis));
    vis[0] = 1;
  }
  int main() {
    int zt, i, j, flat = 0, iz = 0, ip = 0, l, r, ma = 0, mi, x, y, k;
    ios::sync_with_stdio(false);
    cin >> n;
    for (i = 1; i <= n; i++) cin >> a[i];
    init();
    for (i = 1; i <= n; i++) {
      id = a[i];
      for (j = 0; j < p[id].size(); j++) {
        x = p[id][j] ^ id;
        vis[x] = 1;
        for (l = id + 1; l < 520; l++) {
          if (!pis[l][x]) {
            p[l].push_back(x);
            pis[l][x] = 1;
          }
        }
      }
      p[a[i]].clear();
    }
    for (i = 0; i < 8192; i++)
      if (vis[i]) ans.push_back(i);
    cout << ans.size() << endl;
    for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
    return 0;
  }
  