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;
  long long power(long long a, long long b) {
    if (b == 0) {
      return 1;
    }
    long long res = 1;
    while (b > 0) {
      if (b & 1) {
        res = (res * a) % 1000000007;
      }
      a = (a * a) % 1000000007;
      b = b / 2;
    }
    return res;
  }
  long long gcd(long long a, long long b) {
    if (a == 0 || b == 0) {
      return a + b;
    }
    return gcd(b, a % b);
  }
  int32_t main() {
    long long n;
    cin >> n;
    vector<long long> v(n);
    for (long long i = 0; i < n; i++) {
      cin >> v[i];
    }
    long long MX = 1 << 10;
    vector<long long> ans(MX, -1);
    ans[0] = 0;
    long long tmp;
    long long count = 0;
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < MX; j++) {
        if (ans[j] != -1 && v[i] > ans[j]) {
          tmp = j ^ v[i];
          if (ans[tmp] == -1) {
            ans[tmp] = v[i];
            count++;
          } else {
            ans[tmp] = min(ans[tmp], v[i]);
          }
        }
      }
      if (ans[v[i]] == -1) {
        count++;
        ans[v[i]] = v[i];
      } else {
        ans[v[i]] = min(ans[v[i]], v[i]);
      }
    }
    count++;
    cout << count << '\n';
    for (long long i = 0; i < MX; i++) {
      if (ans[i] != -1) {
        cout << i << " ";
      }
    }
    cout << '\n';
  }
  