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;
  namespace IO_BUFF {
  const int BS = (1 << 24) + 5;
  char Buffer[BS], *HD, *TL;
  inline int gc() {
    if (HD == TL) TL = (HD = Buffer) + fread(Buffer, 1, BS, stdin);
    return (HD == TL) ? EOF : *HD++;
  }
  inline int inn() {
    int x, ch;
    while ((ch = gc()) < '0' || ch > '9')
      ;
    x = ch ^ '0';
    while ((ch = gc()) >= '0' && ch <= '9') x = (x << 1) + (x << 3) + (ch ^ '0');
    return x;
  }
  char ss[19999999], tt[20];
  int ssl, ttl;
  inline int print(int x) {
    if (x < 0) ss[++ssl] = '-', x = (-x);
    if (!x) ss[++ssl] = '0';
    for (ttl = 0; x; x /= 10) tt[++ttl] = char(x % 10 + '0');
    for (; ttl; ttl--) ss[++ssl] = tt[ttl];
    return ss[++ssl] = '\n';
  }
  inline int Flush() {
    return fwrite(ss + 1, sizeof(char), ssl, stdout), ssl = 0, 0;
  }
  int read() {
    char c = getchar();
    int x = 1;
    int s = 0;
    while (c < '0' || c > '9') {
      if (c == '-') x = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      s = s * 10 + c - '0';
      c = getchar();
    }
    return s * x;
  }
  }  // namespace IO_BUFF
  using namespace IO_BUFF;
  const int N = 1e5 + 5;
  const int M = 513;
  int f[N][M], a[N], n, m, st[N];
  int main() {
    n = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    m = 512;
    memset(f, 10, sizeof(f));
    for (int i = 1; i <= n; i++) {
      for (int k = 0; k < m; k++) f[i][k] = f[i - 1][k];
      f[i][a[i]] = min(f[i][a[i]], a[i]);
      for (int k = 0; k < m; k++) {
        if (f[i - 1][k] < a[i]) f[i][a[i] ^ k] = min(f[i][a[i] ^ k], a[i]);
      }
    }
    int top = 0;
    st[++top] = 0;
    for (int i = 1; i < m; i++) {
      if (f[n][i] < 1e8) {
        st[++top] = i;
      }
    }
    cout << top << endl;
    for (int i = 1; i <= top; i++) {
      cout << st[i] << " ";
    }
    return 0;
  }
  