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 n, ans, tot, tmp, rk[1000005], a[1000005], vis[(1 << 13)], X[(1 << 13)],
      Y[(1 << 13)], Z[(1 << 13)], ok[(1 << 13)];
  inline bool cmp(int x, int y) {
    if (a[x] ^ a[y]) return a[x] < a[y];
    return x < y;
  }
  struct FastIO {
    static const int S = 1048576;
    char buf[S], *L, *R;
    int stk[20], Top;
    ~FastIO() { clear(); }
    inline char nc() {
      return L == R && (R = (L = buf) + fread(buf, 1, S, stdin), L == R) ? EOF
                                                                         : *L++;
    }
    inline void clear() {
      fwrite(buf, 1, Top, stdout);
      Top = 0;
    }
    inline void pc(char ch) {
      Top == S && (clear(), 0);
      buf[Top++] = ch;
    }
    inline void endl() { pc('\n'); }
    FastIO& operator>>(char& ch) {
      while (ch = nc(), ch == ' ' || ch == '\n' || ch == '\r')
        ;
      return *this;
    }
    template <typename T>
    FastIO& operator>>(T& ret) {
      ret = 0;
      int f = 1;
      char ch = nc();
      while (ch < '0' || ch > '9') {
        if (ch == '-') f = -f;
        ch = nc();
      }
      while (ch >= '0' && ch <= '9') {
        ret = ret * 10 + ch - '0';
        ch = nc();
      }
      ret *= f;
      return *this;
    }
    FastIO& operator>>(char* s) {
      int Len = 0;
      char ch = nc();
      while (ch != '\n') {
        *(s + Len) = ch;
        Len++;
        ch = nc();
      }
    }
    template <typename T>
    FastIO& operator<<(T x) {
      if (x < 0) {
        pc('-');
        x = -x;
      }
      do {
        stk[++stk[0]] = x % 10;
        x /= 10;
      } while (x);
      while (stk[0]) pc('0' + stk[stk[0]--]);
      return *this;
    }
    FastIO& operator<<(char ch) {
      pc(ch);
      return *this;
    }
    FastIO& operator<<(string str) {
      int Len = str.size() - 1;
      for (stk[0] = 0; Len >= 0; Len--) stk[++stk[0]] = str[Len];
      while (stk[0]) pc(stk[stk[0]--]);
      return *this;
    }
  } fin, fout;
  int main() {
    fin >> n;
    for (register int i = 1; i <= n; i++) rk[i] = i, fin >> a[i];
    sort(rk + 1, rk + n + 1, cmp);
    for (register int i = 1; i <= n; i++) {
      int j = i, x = a[rk[i]], tt = 0;
      while (j <= n && x == a[rk[j]]) j++;
      j--;
      if (!vis[x] || vis[x] > rk[i]) vis[x] = rk[i], Y[++tt] = x;
      tmp++;
      for (register int k = 1; k <= tot; k++) {
        while (i <= j && vis[X[k]] >= rk[i]) i++;
        if (i > j) break;
        int y = X[k] ^ x;
        if (vis[y] > rk[i] || !vis[y]) ok[y] = tmp, vis[y] = rk[i], Y[++tt] = y;
      }
      int l = 1, r = 1, T = tot;
      tot = 0;
      while (l <= tt && r <= T) {
        if (ok[X[r]] == tmp) {
          r++;
          continue;
        }
        if (vis[Y[l]] < vis[X[r]])
          Z[++tot] = Y[l], l++;
        else
          Z[++tot] = X[r], r++;
      }
      while (l <= tt) Z[++tot] = Y[l], l++;
      while (r <= T) {
        if (ok[X[r]] == tmp) {
          r++;
          continue;
        }
        Z[++tot] = X[r], r++;
      }
      for (register int k = 1; k <= tot; k++) X[k] = Z[k];
      i = j;
    }
    vis[0] = 1;
    for (register int i = 0; i < (1 << 13); i++)
      if (vis[i]) ans++;
    fout << ans << '\n';
    for (register int i = 0; i < (1 << 13); i++)
      if (vis[i]) fout << i << ' ';
    return 0;
  }
  