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 ld = long double;
  const int MOD = 1e9 + 7;
  int binarySearch(int arr[], int l, int r, int x) {
    if (r >= l) {
      int mid = l + (r - l) / 2;
      if (arr[mid] == x) return mid;
      if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
      return binarySearch(arr, mid + 1, r, x);
    } else {
      return -1;
    }
  }
  ll power(ll x, ll y) {
    ll res = 1;
    while (y > 0) {
      if (y & 1) res = (res * x);
      y = y >> 1;
      x = (x * x);
    }
    return res;
  }
  ll gcd(ll a, ll b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  const int N = 100000;
  int arr[N];
  int mn[512];
  void solve() {
    int n;
    cin >> n;
    for (int i = (0); i < (n); i++) {
      cin >> arr[i];
    }
    mn[0] = -1;
    for (int i = (0); i < (n); i++) {
      for (int j = (0); j < (512); j++) {
        if (mn[j] != MOD && arr[i] > mn[j]) {
          int y = arr[i] ^ j;
          mn[y] = min(mn[y], arr[i]);
        }
      }
    }
    int ans = 0;
    vector<int> val;
    for (int i = (0); i < (512); i++) {
      if (mn[i] != MOD) {
        ans++;
        val.push_back(i);
      }
    }
    cout << ans << '\n';
    for (int i = (0); i < (ans); i++) {
      cout << val[i] << " ";
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    for (int i = (0); i < (512); i++) {
      mn[i] = MOD;
    }
    while (t--) {
      solve();
    }
    return 0;
  }
  