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 ui = unsigned int;
  const double pi = acos(-1.0);
  const int dx[] = {0, 1, 0, -1};
  const int dy[] = {1, 0, -1, 0};
  const int limit = 5e5;
  int read() {
    int s = 0;
    char c = getchar(), lc = '+';
    while (c < '0' || '9' < c) lc = c, c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
    return lc == '-' ? -s : s;
  }
  void write(long long x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10)
      putchar(x + '0');
    else
      write(x / 10), putchar(x % 10 + '0');
  }
  void print(long long x, char c = '\n') {
    write(x);
    putchar(c);
  }
  const int mod = 998244353;
  const int M = 20 + 5;
  const int S = (1 << 20) + 5;
  const int N = 1e5 + 5;
  int n, m, k, _;
  int a[N];
  int dp[600];
  signed main() {
    while (~scanf("%d", &n)) {
      for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
      memset(dp, 0x7f7f7f7f, sizeof(dp));
      dp[0] = 0;
      for (int i = 1; i <= n; i++) {
        for (int j = 0; j < 512; j++) {
          if (a[i] > dp[j]) dp[j ^ a[i]] = min(dp[j ^ a[i]], a[i]);
        }
      }
      vector<int> ans;
      for (int i = 0; i < 512; i++) {
        if (dp[i] != 0x7f7f7f7f) ans.push_back(i);
      }
      printf("%d\n", (ans.size()));
      for (auto it : ans) printf("%d ", it);
    }
    return 0;
  }
  