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 MOD = 1e9 + 7;
  int power(int a, int b) {
    int res = 1;
    a = a % MOD;
    while (b > 0) {
      if (b & 1) {
        res = (res * a) % MOD;
        b--;
      }
      a = (a * a) % MOD;
      b >>= 1;
    }
    return res;
  }
  int fermat_inv(int y) { return power(y, MOD - 2); }
  int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
  int min(int a, int b) { return (a > b) ? b : a; }
  int max(int a, int b) { return (a > b) ? a : b; }
  struct node {
    bitset<515> x;
    node() {
      x.reset();
      x[0] = 1;
    }
    node operator+(const node& other) {
      node sum = node();
      sum.x = (x | other.x);
      return sum;
    }
  };
  struct SegTree {
    vector<node> t;
    int n;
    SegTree(int n) {
      this->n = n;
      t.assign(4 * n, node());
    }
    node sum(int v, int tl, int tr, int l, int r) {
      if (l > r) return node();
      if (l == tl && r == tr) return t[v];
      int tm = (tl + tr) / 2;
      return sum(v * 2, tl, tm, l, min(r, tm)) +
             sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
    }
    bitset<515> sum(int l, int r) {
      l++, r++;
      return sum(1, 1, n, l, r).x;
    }
    void update(int v, int tl, int tr, int pos, bitset<515> new_val) {
      if (tl == tr)
        t[v].x = new_val;
      else {
        int tm = (tl + tr) / 2;
        if (pos <= tm)
          update(2 * v, tl, tm, pos, new_val);
        else
          update(2 * v + 1, tm + 1, tr, pos, new_val);
        t[v] = t[2 * v] + t[2 * v + 1];
      }
    }
    void update(int idx, bitset<515> a) {
      idx++;
      update(1, 1, n, idx, a);
    }
  };
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int te = 1;
    while (te--) {
      int n;
      cin >> n;
      int a[n];
      for (int& i : a) cin >> i;
      vector<int> ans;
      SegTree ST(515 + 5);
      for (int i = 0; i < n; i++) {
        bitset<515> curr, pre;
        pre = ST.sum(0, a[i] - 1);
        curr = pre;
        for (int j = 0; j < 515; j++) {
          if (pre[j]) curr[j ^ a[i]] = 1;
        }
        ST.update(a[i], curr);
      }
      for (int i = 0; i <= 512; i++) {
        if (ST.sum(0, 515)[i]) ans.push_back(i);
      }
      sort((ans).begin(), (ans).end());
      cout << ans.size() << '\n';
      for (int i : ans) cout << i << ' ';
      cout << '\n';
    }
  }
  