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 ull = unsigned long long;
  const ll mod = 1000000007;
  const ll inf = 1e9;
  ll rng(ll l, ll u) { return l + (ll)rand() % (u - l + 1); }
  ll bpow(ll a, int n) {
    return n ? n & 1 ? a * bpow(a, n - 1) : bpow(a * a, n >> 1) : 1;
  }
  vector<int> stupid(int n, vector<int> a) {
    vector<int> ans = {0};
    set<int> s = {0};
    for (int i = 1; i < pow(2, n); ++i) {
      vector<int> m;
      int mask = i, j = n - 1;
      while (mask) {
        if (mask % 2) {
          m.push_back(j);
        }
        mask /= 2;
        j--;
      }
      reverse(m.begin(), m.end());
      int x = a[m[0]];
      bool flag = 1;
      for (int j = 1; j < m.size(); ++j) {
        if (a[m[j]] <= a[m[j - 1]]) {
          flag = 0;
          break;
        }
        x ^= a[m[j]];
      }
      if (flag && !s.count(x)) {
        ans.push_back(x);
        s.insert(x);
      }
    }
    sort(ans.begin(), ans.end());
    return ans;
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    vector<int> ans = {0}, used(513, 0), m(513, 0);
    used[0] = true;
    for (int i = 0; i < n; ++i) {
      int sz = ans.size();
      for (int j = 0; j < sz; ++j) {
        int res = ans[j];
        if (m[res] < a[i]) {
          int new_res = res ^ a[i];
          if (m[new_res]) {
            m[new_res] = min(a[i], m[new_res]);
          } else {
            if (new_res) m[new_res] = a[i];
          }
          if (!used[new_res]) {
            ans.push_back(new_res);
            used[new_res] = true;
          }
        }
      }
    }
    sort(ans.begin(), ans.end());
    cout << ans.size() << endl;
    for (int i = 0; i < ans.size(); ++i) {
      cout << ans[i] << ' ';
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    for (int i = 0; i < t; ++i) {
      solve();
      cout << endl;
    }
  }
  