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("Ofast")
  #pragma GCC target("avx2")
  using namespace std;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long pow2(long long a, long long b, long long m) {
    long long ans = 1;
    while (b > 0) {
      if (b & 1) {
        ans = (ans * a) % m;
      }
      a = (a * a) % m;
      b >>= 1;
    }
    ans = (ans + m) % m;
    return ans;
  }
  long long pow2(long long a, long long b) {
    long long ans = 1;
    while (b > 0) {
      if (b & 1) {
        ans = (ans * a);
      }
      a = (a * a);
      b >>= 1;
    }
    return ans;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> pos(1024, INT_MAX);
    pos[0] = 0;
    for (long long i = 0; i < n; i++) {
      long long a;
      cin >> a;
      for (long long j = 0; j < 1024; j++) {
        if (pos[j] < a) pos[j ^ a] = min(pos[j ^ a], a);
      }
    }
    vector<long long> v;
    for (long long i = 0; i < 1024; i++) {
      if (pos[i] != INT_MAX) v.push_back(i);
    }
    cout << v.size() << "\n";
    for (auto x : v) cout << x << " ";
    cout << "\n";
  }
  int32_t main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(NULL);
    std::cout.tie(NULL);
    long long t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  