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;
  void show(vector<long long> &a) {
    for (long long e : a) {
      if (e == (long long)(1e18 + 5))
        cout << "INF ";
      else
        cout << e << " ";
    }
    cout << "\n";
  }
  void show(vector<int> &a) {
    for (int e : a) {
      if (e == (long long)(1e9 + 7))
        cout << "MOD ";
      else
        cout << e << " ";
    }
    cout << "\n";
  }
  template <typename T>
  void enter(vector<T> &a) {
    for (T &e : a) cin >> e;
  }
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int first = 0;
    cerr << '{';
    for (auto &i : x) cerr << (first++ ? "," : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  void JUDGE() {}
  long long inline mo(long long a) { return a % (long long)(1e9 + 7); }
  long long po(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y >>= 1;
      x = (x * x) % p;
    }
    return res % p;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n;
    vector<int> a(n);
    enter(a);
    int li = (1 << 9);
    vector<int> val(li, (long long)(1e9 + 7));
    val[0] = -1;
    for (int i = 0; i < n; i++) {
      vector<int> new_val = val;
      for (int mask = 0; mask < li; mask++) {
        if (val[mask] < a[i]) {
          int next = (mask ^ a[i]);
          new_val[next] = min(new_val[next], a[i]);
        }
      }
      val = new_val;
    }
    vector<int> ans;
    for (int i = 0; i < li; i++) {
      if (val[i] < (long long)(1e9 + 7)) ans.push_back(i);
    }
    cout << (int)ans.size() << "\n";
    for (int e : ans) cout << e << " ";
    cout << "\n";
    return 0;
  }
  