#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vvvi = vector<vector<vector<int>>>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vvvl = vector<vector<vector<ll>>>;
using vs = vector<string>;
using vb = vector<bool>;
using vvb = vector<vector<bool>>;
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline void prn(vector<T>& v) {
  int n = (int)v.size();
  for (int i = (0); i < ((n)); i++) cout << v[i] << ' ';
}
template <class T>
inline void printv(vector<T>& v) {
  int n = (int)v.size();
  for (int i = (0); i < ((n)); i++) cout << v[i] << (i == n - 1 ? '\n' : ' ');
}
template <class T>
inline void printvv(vector<vector<T>>& v) {
  for (auto u : v) printv(u);
}
template <class T>
inline void printvm(vector<T>& v) {
  int n = (int)v.size();
  for (int i = (0); i < ((n)); i++)
    cout << v[i].val() << (i == n - 1 ? '\n' : ' ');
}
template <class T>
inline void printvvm(vector<vector<T>>& v) {
  for (auto u : v) printvm(u);
}
template <class T>
inline void printlnv(vector<T>& v) {
  int n = (int)v.size();
  for (int i = (0); i < ((n)); i++) cout << v[i] << '\n';
}
const int MOD = 1000000007;
const int INF = 1000000001;
const ll LINF = 1000000001000000001LL;
void solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(numeric_limits<double>::max_digits10);
  solve();
  return 0;
}
void solve() {
  int n;
  cin >> n;
  vi a(n);
  for (int i = (0); i < ((n)); i++) cin >> a[i];
  vvb dp(n + 1, vb(513, false));
  vi dp2(513, INF);
  dp[0][0] = true;
  dp2[0] = 0;
  for (int i = (0); i < ((n)); i++) {
    for (int j = (0); j < ((513)); j++) {
      if (dp[i][j]) {
        dp[i + 1][j] = true;
        if (dp2[j] < a[i]) {
          dp[i + 1][j ^ a[i]] = true;
          chmin(dp2[j ^ a[i]], a[i]);
        }
      }
    }
  }
  vi ans;
  int cnt = 0;
  for (int i = (0); i < ((513)); i++)
    if (dp[n][i]) {
      ans.emplace_back(i);
      cnt++;
    }
  cout << cnt << '\n';
  printv(ans);
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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] 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vvvi = vector<vector<vector<int>>>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vvvl = vector<vector<vector<ll>>>;
using vs = vector<string>;
using vb = vector<bool>;
using vvb = vector<vector<bool>>;
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline void prn(vector<T>& v) {
  int n = (int)v.size();
  for (int i = (0); i < ((n)); i++) cout << v[i] << ' ';
}
template <class T>
inline void printv(vector<T>& v) {
  int n = (int)v.size();
  for (int i = (0); i < ((n)); i++) cout << v[i] << (i == n - 1 ? '\n' : ' ');
}
template <class T>
inline void printvv(vector<vector<T>>& v) {
  for (auto u : v) printv(u);
}
template <class T>
inline void printvm(vector<T>& v) {
  int n = (int)v.size();
  for (int i = (0); i < ((n)); i++)
    cout << v[i].val() << (i == n - 1 ? '\n' : ' ');
}
template <class T>
inline void printvvm(vector<vector<T>>& v) {
  for (auto u : v) printvm(u);
}
template <class T>
inline void printlnv(vector<T>& v) {
  int n = (int)v.size();
  for (int i = (0); i < ((n)); i++) cout << v[i] << '\n';
}
const int MOD = 1000000007;
const int INF = 1000000001;
const ll LINF = 1000000001000000001LL;
void solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(numeric_limits<double>::max_digits10);
  solve();
  return 0;
}
void solve() {
  int n;
  cin >> n;
  vi a(n);
  for (int i = (0); i < ((n)); i++) cin >> a[i];
  vvb dp(n + 1, vb(513, false));
  vi dp2(513, INF);
  dp[0][0] = true;
  dp2[0] = 0;
  for (int i = (0); i < ((n)); i++) {
    for (int j = (0); j < ((513)); j++) {
      if (dp[i][j]) {
        dp[i + 1][j] = true;
        if (dp2[j] < a[i]) {
          dp[i + 1][j ^ a[i]] = true;
          chmin(dp2[j ^ a[i]], a[i]);
        }
      }
    }
  }
  vi ans;
  int cnt = 0;
  for (int i = (0); i < ((513)); i++)
    if (dp[n][i]) {
      ans.emplace_back(i);
      cnt++;
    }
  cout << cnt << '\n';
  printv(ans);
}

```