#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  long long arr[n];
  bitset<1024> mp[501];
  bitset<1024> ans;
  ans.set(0);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    mp[arr[i]].set(arr[i]);
    ans.set(arr[i]);
    for (long long j = arr[i] - 1; j >= 0; j--) {
      mp[arr[i]] |= mp[j];
    }
    for (long long j = 0; j < 1024; j++) {
      if (mp[arr[i]].test(j)) {
        mp[arr[i]].set(arr[i] ^ j);
      }
    }
    ans |= mp[arr[i]];
  }
  cout << ans.count() << "\n";
  for (long long i = 0; i < 1024; i++) {
    if (ans.test(i)) {
      cout << i << " ";
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  solve();
  return 0;
}

### Prompt

Your task is to create 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() {
  long long n;
  cin >> n;
  long long arr[n];
  bitset<1024> mp[501];
  bitset<1024> ans;
  ans.set(0);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    mp[arr[i]].set(arr[i]);
    ans.set(arr[i]);
    for (long long j = arr[i] - 1; j >= 0; j--) {
      mp[arr[i]] |= mp[j];
    }
    for (long long j = 0; j < 1024; j++) {
      if (mp[arr[i]].test(j)) {
        mp[arr[i]].set(arr[i] ^ j);
      }
    }
    ans |= mp[arr[i]];
  }
  cout << ans.count() << "\n";
  for (long long i = 0; i < 1024; i++) {
    if (ans.test(i)) {
      cout << i << " ";
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  solve();
  return 0;
}

```