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>
  #pragma GCC optimize("O3")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  const long long INF = 1e18 + 18;
  const long long MOD = 1000000007;
  template <class T>
  bool chmax(T &a, const T &b) {
    if (a < b) {
      a = b;
      return true;
    }
    return false;
  }
  template <class T>
  bool chmin(T &a, const T &b) {
    if (b < a) {
      a = b;
      return true;
    }
    return false;
  }
  template <typename T1, typename T2>
  ostream &operator<<(ostream &s, const pair<T1, T2> &p) {
    return s << "(" << p.first << ", " << p.second << ")";
  }
  template <typename T>
  istream &operator>>(istream &i, vector<T> &v) {
    for (long long j = (0); j < (v.size()); j++) i >> v[j];
    return i;
  }
  template <typename T>
  ostream &operator<<(ostream &s, const vector<T> &v) {
    int len = v.size();
    for (int i = 0; i < len; ++i) {
      s << v[i];
      if (i < len - 1) s << "	";
    }
    return s;
  }
  template <typename T>
  ostream &operator<<(ostream &s, const vector<vector<T>> &vv) {
    s << '\n';
    int len = vv.size();
    for (int i = 0; i < len; ++i) {
      s << vv[i] << '\n';
    }
    return s;
  }
  int solve() {
    long long n;
    cin >> n;
    vector<long long> a(n);
    cin >> a;
    long long mx = 701;
    vector<long long> dp(mx + 1, INF);
    dp[0] = 0;
    for (long long i = (0); i < (n); i++) {
      vector<long long> pre(mx + 1, INF);
      swap(pre, dp);
      for (long long j = (0); j < (mx + 1); j++) {
        long long v = a[i];
        if (pre[j] < v) {
          if ((j ^ v) <= 701) {
            chmin(dp[j ^ v], v);
          }
        }
        chmin(dp[j], pre[j]);
      }
    }
    vector<long long> ans;
    for (long long i = (0); i < (mx); i++) {
      if (dp[i] != INF) {
        ans.push_back(i);
      }
    }
    cout << ans.size() << '\n';
    for (long long i = (0); i < (ans.size()); i++) {
      cout << ans[i] << " ";
    }
    cout << '\n';
    return 0;
  }
  int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << std::setprecision(10);
    solve();
  }
  