#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long ap[100005], bp[550][550];
vector<int> ve;
int main() {
  int n;
  for (int i = 0; i <= 500; i++) bp[i][0] = 1;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ap[i];
    if (ap[i] == 0) continue;
    for (int j = 0; j <= 512; j++) {
      if (!bp[ap[i]][j]) continue;
      if (bp[ap[i]][j ^ ap[i]]) continue;
      bp[ap[i]][j ^ ap[i]] = 1;
      for (int k = ap[i] + 1; k <= 512; k++) {
        bp[k][j ^ ap[i]] = 1;
      }
    }
  }
  int fl = 0;
  for (int i = 0; i <= 512; i++) {
    if (bp[500][i] == 1) {
      ve.push_back(i);
    }
  }
  int len = ve.size();
  cout << len << endl;
  for (int i = 0; i < len; i++) {
    if (i != 0) cout << " ";
    cout << ve[i];
  }
  cout << endl;
}

### Prompt

Create a solution in Cpp for 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")
using namespace std;
long long ap[100005], bp[550][550];
vector<int> ve;
int main() {
  int n;
  for (int i = 0; i <= 500; i++) bp[i][0] = 1;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> ap[i];
    if (ap[i] == 0) continue;
    for (int j = 0; j <= 512; j++) {
      if (!bp[ap[i]][j]) continue;
      if (bp[ap[i]][j ^ ap[i]]) continue;
      bp[ap[i]][j ^ ap[i]] = 1;
      for (int k = ap[i] + 1; k <= 512; k++) {
        bp[k][j ^ ap[i]] = 1;
      }
    }
  }
  int fl = 0;
  for (int i = 0; i <= 512; i++) {
    if (bp[500][i] == 1) {
      ve.push_back(i);
    }
  }
  int len = ve.size();
  cout << len << endl;
  for (int i = 0; i < len; i++) {
    if (i != 0) cout << " ";
    cout << ve[i];
  }
  cout << endl;
}

```