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;
  int read() {
    int qrx = 0, qry = 1;
    char qrc = getchar();
    while (qrc < '0' || qrc > '9') {
      if (qrc == '-') {
        qry = -1, qrc = getchar();
      } else
        qrc = getchar();
    }
    while (qrc >= '0' && qrc <= '9') qrx = qrx * 10 + qrc - 48, qrc = getchar();
    return qrx * qry;
  }
  const int N = 1e6 + 7, M = 2000 + 7, Mod = 998244353, INF = 1e9 + 7;
  int n, a[N], f[M][2], ans, pr[M];
  int main() {
    n = read();
    int p = 0;
    for (int j = 0; j < M; j++) f[j][0] = INF;
    f[0][0] = -1;
    for (int i = 1; i <= n; i++) a[i] = read();
    for (int i = 1; i <= n; i++)
      if (a[i]) {
        p ^= 1;
        for (int j = 0; j < M; j++) f[j][p] = f[j][p ^ 1];
        for (int j = 0; j < M; j++)
          if (f[j][p ^ 1] < a[i]) f[j ^ a[i]][p] = min(f[j ^ a[i]][p], a[i]);
      }
    for (int j = 0; j < M; j++)
      if (f[j][p] < INF) ++ans;
    printf("%d\n", ans);
    for (int j = 0; j < M; j++)
      if (f[j][p] < INF) printf("%d ", j);
    printf("\n");
    return 0;
  }
  