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 long long int modd = 65536;
  const long long 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;
  long long int id = 0, n, m, tot = 0, ttt = 0, mod, a[100010], b[100010],
                c[100010];
  string s;
  vector<long long int> ans;
  void innt() {
    memset(b, 0, sizeof(b));
    memset(c, 0, sizeof(c));
  }
  int main() {
    long long 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;
    innt();
    ans.clear();
    flat = 0;
    for (i = 1; i <= n; i++) {
      cin >> a[i];
      if (a[i] == 0) flat = 1;
    }
    for (i = 1; i <= n; i++) {
      for (j = 0; j < 520; j++) {
        c[j] = b[j];
      }
      for (j = 1; j < 520; j++) {
        if (c[j] && c[j] < a[i]) {
          x = a[i] ^ j;
          if (c[x] != 0)
            b[x] = min(a[i], c[x]);
          else
            b[x] = a[i];
        }
      }
      if (c[a[i]]) {
        flat = 1;
        b[a[i]] = min(c[a[i]], a[i]);
      } else
        b[a[i]] = a[i];
    }
    ans.push_back(0);
    for (i = 1; i < 520; i++) {
      if (b[i]) ans.push_back(i);
    }
    cout << ans.size() << endl;
    sort(ans.begin(), ans.end());
    for (i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
    return 0;
  }
  