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;
  const int N = 1e5 + 10;
  int a[N], ans[N], f[700][700];
  int lowbit(int x) { return x & (-x); }
  int n, c[1100][1100];
  int query(int u, int v) {
    int ans = 0;
    while (u > 0) {
      ans += c[v][u];
      u -= lowbit(u);
    }
    return ans;
  }
  void add(int u, int v) {
    while (u <= 1025) {
      c[v][u]++;
      u += lowbit(u);
    }
  }
  int main() {
    add(1, 0);
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      for (int j = 0; j <= 512; j++) {
        if (query(a[i], j) > 0) add(a[i] + 1, a[i] ^ j);
      }
      add(a[i] + 1, a[i]);
    }
    vector<int> ans;
    for (int i = 0; i <= 1025; i++)
      if (query(1025, i) > 0) ans.push_back(i);
    printf("%d\n", ans.size());
    for (auto x : ans) printf("%d ", x);
    puts("");
    return 0;
  }
  