#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops, no-stack-protector")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using vb = vector<bool>;
using vi = vector<int>;
using vl = vector<long long>;
using vpii = vector<pair<int, int>>;
using vvb = vector<vector<bool>>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<long long>>;
using vvpii = vector<vector<pair<int, int>>>;
using mii = map<int, int>;
using umii = unordered_map<int, int>;
using seti = set<int>;
using useti = unordered_set<int>;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vb res(8200);
  vvi store(8200, vi(1, 0));
  vvb seen(8200, vb(8200));
  res[0] = 1;
  for (int i = 0; i < (8200); i++) seen[i][0] = true;
  for (int i = 0; i < (n); i++) {
    int x;
    cin >> x;
    for (auto d : store[x]) {
      int y = d ^ x;
      res[y] = true;
      for (int j = x + 1; j < 8200; j++)
        if (!seen[j][y]) {
          seen[j][y] = true;
          store[j].push_back(y);
        } else
          break;
    }
    store[x].clear();
  }
  int ans = 0;
  for (int i = 0; i < (8200); i++) ans += res[i];
  cout << ans << "\n";
  for (int i = 0; i < (8200); i++)
    if (res[i]) cout << i << " ";
  cout << "\n";
}

### 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops, no-stack-protector")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using vb = vector<bool>;
using vi = vector<int>;
using vl = vector<long long>;
using vpii = vector<pair<int, int>>;
using vvb = vector<vector<bool>>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<long long>>;
using vvpii = vector<vector<pair<int, int>>>;
using mii = map<int, int>;
using umii = unordered_map<int, int>;
using seti = set<int>;
using useti = unordered_set<int>;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vb res(8200);
  vvi store(8200, vi(1, 0));
  vvb seen(8200, vb(8200));
  res[0] = 1;
  for (int i = 0; i < (8200); i++) seen[i][0] = true;
  for (int i = 0; i < (n); i++) {
    int x;
    cin >> x;
    for (auto d : store[x]) {
      int y = d ^ x;
      res[y] = true;
      for (int j = x + 1; j < 8200; j++)
        if (!seen[j][y]) {
          seen[j][y] = true;
          store[j].push_back(y);
        } else
          break;
    }
    store[x].clear();
  }
  int ans = 0;
  for (int i = 0; i < (8200); i++) ans += res[i];
  cout << ans << "\n";
  for (int i = 0; i < (8200); i++)
    if (res[i]) cout << i << " ";
  cout << "\n";
}

```