#include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9 + 7;
int n, a[100100], f[(1LL << (9)) + 7];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  memset(f, 0x3f, sizeof f);
  f[0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int mask = 0; mask < (1LL << (9)); mask++)
      if (f[mask] < a[i])
        f[mask ^ a[i]] =
            f[mask ^ a[i]] >= INF ? a[i] : min(f[mask ^ a[i]], a[i]);
  }
  vector<int> res;
  for (int i = 0; i < (1LL << (9)); i++)
    if (f[i] < INF) res.push_back(i);
  cout << (int)res.size() << "\n";
  for (auto p : res) cout << p << " ";
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
const int INF = (int)1e9 + 7;
int n, a[100100], f[(1LL << (9)) + 7];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  memset(f, 0x3f, sizeof f);
  f[0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int mask = 0; mask < (1LL << (9)); mask++)
      if (f[mask] < a[i])
        f[mask ^ a[i]] =
            f[mask ^ a[i]] >= INF ? a[i] : min(f[mask ^ a[i]], a[i]);
  }
  vector<int> res;
  for (int i = 0; i < (1LL << (9)); i++)
    if (f[i] < INF) res.push_back(i);
  cout << (int)res.size() << "\n";
  for (auto p : res) cout << p << " ";
}

```