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;
  using ll = long long;
  using ld = long double;
  using vi = vector<int>;
  using vvi = vector<vector<int>>;
  using vvvi = vector<vector<vector<int>>>;
  using vl = vector<ll>;
  using vvl = vector<vector<ll>>;
  using vvvl = vector<vector<vector<ll>>>;
  using vs = vector<string>;
  using vb = vector<bool>;
  using vvb = vector<vector<bool>>;
  template <class T>
  inline bool chmin(T& a, T b) {
    if (a > b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class T>
  inline bool chmax(T& a, T b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class T>
  inline void prn(vector<T>& v) {
    int n = (int)v.size();
    for (int i = (0); i < ((n)); i++) cout << v[i] << ' ';
  }
  template <class T>
  inline void printv(vector<T>& v) {
    int n = (int)v.size();
    for (int i = (0); i < ((n)); i++) cout << v[i] << (i == n - 1 ? '\n' : ' ');
  }
  template <class T>
  inline void printvv(vector<vector<T>>& v) {
    for (auto u : v) printv(u);
  }
  template <class T>
  inline void printvm(vector<T>& v) {
    int n = (int)v.size();
    for (int i = (0); i < ((n)); i++)
      cout << v[i].val() << (i == n - 1 ? '\n' : ' ');
  }
  template <class T>
  inline void printvvm(vector<vector<T>>& v) {
    for (auto u : v) printvm(u);
  }
  template <class T>
  inline void printlnv(vector<T>& v) {
    int n = (int)v.size();
    for (int i = (0); i < ((n)); i++) cout << v[i] << '\n';
  }
  const int MOD = 1000000007;
  const int INF = 1000000001;
  const ll LINF = 1000000001000000001LL;
  void solve();
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(numeric_limits<double>::max_digits10);
    solve();
    return 0;
  }
  void solve() {
    int n;
    cin >> n;
    vi a(n);
    for (int i = (0); i < ((n)); i++) cin >> a[i];
    vvb dp(n + 1, vb(513, false));
    vi dp2(513, INF);
    dp[0][0] = true;
    dp2[0] = 0;
    for (int i = (0); i < ((n)); i++) {
      for (int j = (0); j < ((513)); j++) {
        if (dp[i][j]) {
          dp[i + 1][j] = true;
          if (dp2[j] < a[i]) {
            dp[i + 1][j ^ a[i]] = true;
            chmin(dp2[j ^ a[i]], a[i]);
          }
        }
      }
    }
    vi ans;
    int cnt = 0;
    for (int i = (0); i < ((513)); i++)
      if (dp[n][i]) {
        ans.emplace_back(i);
        cnt++;
      }
    cout << cnt << '\n';
    printv(ans);
  }
  