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;
  template <class c>
  struct rge {
    c b, e;
  };
  template <class c>
  rge<c> range(c i, c j) {
    return rge<c>{i, j};
  }
  template <class c>
  auto dud(c *x) -> decltype(cerr << *x, 0);
  template <class c>
  char dud(...);
  struct debug {
    template <class c>
    debug &operator<<(const c &) {
      return *this;
    }
  };
  long long __pow(long long x, long long y, long long M) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) res = (res * x) % M;
      y >>= 1;
      x = (x * x) % M;
    }
    return res % M;
  }
  long long __pow(long long x, long long y) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) res = (res * x);
      y >>= 1;
      x = (x * x);
    }
    return res;
  }
  long long mi(long long x, long long M) { return __pow(x, M - 2, M); }
  long long mul(vector<long long> a, long long M) {
    long long res = 1;
    for (auto x : a) res = (res * x) % M;
    return res;
  }
  long long cl(long long a, long long b) {
    if (a % b == 0) return a / b;
    return a / b + 1;
  }
  void umax(long long &a, long long b) { a = max(a, b); }
  void umin(long long &a, long long b) { a = min(a, b); }
  const long long N = 2e5 + 2, inf = 1e18, M = 998244353, LN = 20;
  long long Test, n, m, k, q, a[N];
  bool ok = 1;
  class BIT {
    long long n, VAL;
    vector<long long> bit;
  
   public:
    BIT() {}
    BIT(long long _n, long long val) {
      n = _n;
      VAL = val;
      bit.assign(2 * n + 2, VAL);
    }
    long long todo(long long a, long long b) { return a + b; }
    void update(long long i, long long x) {
      i++;
      for (; i <= n; i += (i & -i)) bit[i] = todo(bit[i], x);
    }
    long long get(long long i) {
      i++;
      long long res = VAL;
      for (; i; i -= (i & -i)) res = todo(res, bit[i]);
      return res;
    }
  };
  void test_case() {
    cin >> n;
    vector<BIT> trees(512);
    for (long long i = 0; i <= 511; i++) trees[i] = BIT(512, 0);
    vector<bool> vis(512);
    vis[0] = 1;
    long long cnt = 1;
    for (long long i = 1, x; i <= n; i++) {
      cin >> x;
      if (!vis[x]) cnt++;
      vis[x] = 1;
      trees[x].update(x, 1);
      for (long long j = 1; j <= 511; j++)
        if (x > 0 && (x ^ j) > 0 && trees[x ^ j].get(x - 1) > 0)
          cnt += !vis[j], vis[j] = 1, trees[j].update(x, 1);
    }
    cout << cnt << '\n';
    for (long long i = 0; i < 512; i++)
      if (vis[i]) cout << i << " ";
    cout << '\n';
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    Test = 1;
    for (long long t = 1; t <= Test; t++) {
      test_case();
    }
    return 0;
  }
  