#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
bool dp[MAXN][601];
int n, a[MAXN], minv[601];
void solve() {
  cin >> n;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (0); i <= (511); ++i) {
    minv[i] = 0x7f7f7f7f;
  }
  dp[0][0] = 1;
  minv[0] = -1;
  for (int i = (1); i <= (n); ++i) {
    for (int j = (0); j <= (511); ++j)
      if ((a[i] ^ j) <= 511) {
        dp[i][j] |= (dp[i - 1][j]);
        dp[i][j] |= (minv[a[i] ^ j] < a[i]);
        if ((minv[a[i] ^ j] < a[i])) minv[j] = min(minv[j], a[i]);
      }
    dp[i][a[i]] = 1;
    minv[a[i]] = min(minv[a[i]], a[i]);
  }
  int cnt = 0;
  for (int i = (0); i <= (511); ++i) cnt += (dp[n][i]);
  cout << cnt << '\n';
  for (int i = (0); i <= (511); ++i)
    if (dp[n][i]) {
      cout << i << ' ';
    }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int _;
  _ = 1;
  while (_--) {
    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;
const int MAXN = 1e5 + 10;
bool dp[MAXN][601];
int n, a[MAXN], minv[601];
void solve() {
  cin >> n;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (0); i <= (511); ++i) {
    minv[i] = 0x7f7f7f7f;
  }
  dp[0][0] = 1;
  minv[0] = -1;
  for (int i = (1); i <= (n); ++i) {
    for (int j = (0); j <= (511); ++j)
      if ((a[i] ^ j) <= 511) {
        dp[i][j] |= (dp[i - 1][j]);
        dp[i][j] |= (minv[a[i] ^ j] < a[i]);
        if ((minv[a[i] ^ j] < a[i])) minv[j] = min(minv[j], a[i]);
      }
    dp[i][a[i]] = 1;
    minv[a[i]] = min(minv[a[i]], a[i]);
  }
  int cnt = 0;
  for (int i = (0); i <= (511); ++i) cnt += (dp[n][i]);
  cout << cnt << '\n';
  for (int i = (0); i <= (511); ++i)
    if (dp[n][i]) {
      cout << i << ' ';
    }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int _;
  _ = 1;
  while (_--) {
    solve();
  }
  return 0;
}

```