#include <bits/stdc++.h>
using namespace std;
void solve() {
  int mx = 500 + 1, mxor = 512;
  int n;
  cin >> n;
  vector<int> v(n + 1);
  vector<vector<int>> idx(mx);
  for (int i = 1; i <= n; ++i) {
    cin >> v[i];
    idx[v[i]].push_back(i);
  }
  vector<int> mi(mxor, n + 1);
  mi[0] = 0;
  vector<vector<int>> dp(mx, vector<int>(mxor, n + 1));
  dp[0][0] = 0;
  for (int i = 1; i < mx; ++i) {
    for (int j = 0; j < mxor; ++j) {
      int c = i ^ j;
      int z = mi[c];
      auto nx = lower_bound(idx[i].begin(), idx[i].end(), z + 1);
      if (nx != idx[i].end()) {
        dp[i][j] = *nx;
      }
    }
    for (int j = 0; j < mxor; ++j) {
      mi[j] = min(mi[j], dp[i][j]);
    }
  }
  vector<int> a;
  for (int j = 0; j < mxor; ++j) {
    for (int i = 0; i < mx; ++i) {
      if (dp[i][j] <= n) {
        a.push_back(j);
        break;
      }
    }
  }
  cout << a.size() << endl;
  for (auto x : a) {
    cout << x << ' ';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}

### Prompt

Please formulate 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;
void solve() {
  int mx = 500 + 1, mxor = 512;
  int n;
  cin >> n;
  vector<int> v(n + 1);
  vector<vector<int>> idx(mx);
  for (int i = 1; i <= n; ++i) {
    cin >> v[i];
    idx[v[i]].push_back(i);
  }
  vector<int> mi(mxor, n + 1);
  mi[0] = 0;
  vector<vector<int>> dp(mx, vector<int>(mxor, n + 1));
  dp[0][0] = 0;
  for (int i = 1; i < mx; ++i) {
    for (int j = 0; j < mxor; ++j) {
      int c = i ^ j;
      int z = mi[c];
      auto nx = lower_bound(idx[i].begin(), idx[i].end(), z + 1);
      if (nx != idx[i].end()) {
        dp[i][j] = *nx;
      }
    }
    for (int j = 0; j < mxor; ++j) {
      mi[j] = min(mi[j], dp[i][j]);
    }
  }
  vector<int> a;
  for (int j = 0; j < mxor; ++j) {
    for (int i = 0; i < mx; ++i) {
      if (dp[i][j] <= n) {
        a.push_back(j);
        break;
      }
    }
  }
  cout << a.size() << endl;
  for (auto x : a) {
    cout << x << ' ';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}

```