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>
  using namespace std;
  const int MAXN = 1e5 + 10;
  const int MAXA = 511;
  bool dp[MAXA + 1];
  int n, a[MAXN], minv[MAXA + 1];
  void solve() {
    cin >> n;
    for (int i = (1); i <= (n); ++i) cin >> a[i];
    for (int i = (0); i <= (511); ++i) {
      minv[i] = 0x7f7f7f7f;
    }
    dp[0] = 1;
    minv[0] = -1;
    for (int i = (1); i <= (n); ++i) {
      for (int j = (0); j <= (511); ++j) {
        dp[j] |= (minv[a[i] ^ j] < a[i]);
        if ((minv[a[i] ^ j] < a[i])) minv[j] = min(minv[j], a[i]);
      }
    }
    int cnt = 0;
    for (int i = (0); i <= (511); ++i) cnt += (dp[i]);
    cout << cnt << '\n';
    for (int i = (0); i <= (511); ++i)
      if (dp[i]) {
        cout << i << ' ';
      }
    cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int _;
    _ = 1;
    while (_--) {
      solve();
    }
    return 0;
  }
  