#include <bits/stdc++.h>
const int mod = 998244353;
const int modd = 65536;
const int base = 233;
const double pi = acos(-1);
const int N = 2e5 + 10;
const int M = 1e9;
const double esp = 1e-6;
using namespace std;
int id = 0, n, m, tot = 0, ttt = 0, a[1000010];
vector<int> p[100010], ans;
bool vis[100010], pis[8192][8192];
void init() {
  for (int i = 0; i <= 8192; i++) p[i].push_back(0);
  memset(vis, 0, sizeof(vis));
  memset(pis, 0, sizeof(pis));
  vis[0] = 1;
}
int main() {
  int zt, i, j, flat = 0, iz = 0, ip = 0, l, r, ma = 0, mi, x, y, k;
  ios::sync_with_stdio(false);
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  init();
  for (i = 1; i <= n; i++) {
    id = a[i];
    for (j = 0; j < p[id].size(); j++) {
      x = p[id][j] ^ id;
      vis[x] = 1;
      for (l = id + 1; l < 520; l++) {
        if (!pis[l][x]) {
          p[l].push_back(x);
          pis[l][x] = 1;
        }
      }
    }
    p[a[i]].clear();
  }
  for (i = 0; i < 8192; i++)
    if (vis[i]) ans.push_back(i);
  cout << ans.size() << endl;
  for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  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>
const int mod = 998244353;
const int modd = 65536;
const int base = 233;
const double pi = acos(-1);
const int N = 2e5 + 10;
const int M = 1e9;
const double esp = 1e-6;
using namespace std;
int id = 0, n, m, tot = 0, ttt = 0, a[1000010];
vector<int> p[100010], ans;
bool vis[100010], pis[8192][8192];
void init() {
  for (int i = 0; i <= 8192; i++) p[i].push_back(0);
  memset(vis, 0, sizeof(vis));
  memset(pis, 0, sizeof(pis));
  vis[0] = 1;
}
int main() {
  int zt, i, j, flat = 0, iz = 0, ip = 0, l, r, ma = 0, mi, x, y, k;
  ios::sync_with_stdio(false);
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  init();
  for (i = 1; i <= n; i++) {
    id = a[i];
    for (j = 0; j < p[id].size(); j++) {
      x = p[id][j] ^ id;
      vis[x] = 1;
      for (l = id + 1; l < 520; l++) {
        if (!pis[l][x]) {
          p[l].push_back(x);
          pis[l][x] = 1;
        }
      }
    }
    p[a[i]].clear();
  }
  for (i = 0; i < 8192; i++)
    if (vis[i]) ans.push_back(i);
  cout << ans.size() << endl;
  for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  return 0;
}

```