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;
  long long GCD(long long a, long long b) {
    if (b > a) {
      swap(a, b);
    }
    if (b == 0) {
      return a;
    }
    return GCD(b, a % b);
  }
  long long mul_mod(long long a, long long b, long long m) {
    a %= m;
    b %= m;
    return ((((a * b) % m) + m) % m);
  }
  long long power(long long base, long long n, long long m) {
    long long res = 1;
    while (n) {
      if (n & 1) res = mul_mod(res, base, m);
      base = mul_mod(base, base, m);
      n = n >> 1;
    }
    return res;
  }
  long long ncr(long long x, long long y) {
    long long ans = 1;
    y = (y > x - y ? x - y : y);
    for (long long i = 0; i < y; i++) {
      ans = (ans * (x - i)) % 1000000007;
      ans = (ans * power(i + 1, 1000000007 - 2, 1000000007)) % 1000000007;
      ans = ans % 1000000007;
    }
    return ans;
  }
  bool isPrime(long long n) {
    if (n == 1 or n == 0) return false;
    if (n == 2 or n == 3) return true;
    if (n % 2 == 0 or n % 3 == 0) return false;
    for (long long i = 5; i * i <= n; i += 6) {
      if (n % i == 0 or n % (i + 2) == 0) return false;
    }
    return true;
  }
  long long add_mod(long long a, long long b, long long m) {
    a % m;
    b % m;
    return ((((a + b) % m) + m) % m);
  }
  long long sub_mod(long long a, long long b, long long m) {
    a % m;
    b % m;
    return ((((a - b) % m) + m) % m);
  }
  vector<long long> prmFactors(long long n) {
    vector<long long> res;
    for (long long i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        while (n % i == 0) n /= i;
        res.push_back(i);
      }
    }
    if (n > 1) res.push_back(n);
    return res;
  }
  void run() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ;
  }
  void solve() {
    long long n;
    cin >> n;
    long long a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<long long> mn(513, 1e9);
    vector<long long> ans;
    ans.push_back(0);
    mn[0] = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 513; j++) {
        if (mn[j] < a[i]) {
          mn[j ^ a[i]] = min(mn[j ^ a[i]], a[i]);
        }
      }
    }
    for (int i = 1; i < 513; i++)
      if (mn[i] != 1e9) ans.push_back(i);
    cout << ans.size() << endl;
    for (auto x : ans) cout << x << " ";
    cout << endl;
  }
  int main() {
    run();
    int t = 1;
    while (t--) {
      solve();
    }
    return 0LL;
  }
  