#include <bits/stdc++.h>
using namespace std;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
int T;
int N, a[100005];
bool sums[600];
int p[600];
int main() {
  fast();
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> a[i];
  sums[0] = true;
  for (int i = 1; i <= N; i++) {
    for (int s = 0; s <= 512; s++)
      if (sums[s] != 0 && p[s] < a[i] && sums[(s ^ a[i])] == 0) {
        sums[(s ^ a[i])] = true;
        p[(s ^ a[i])] = a[i];
      } else if (sums[s] != 0 && p[s] < a[i] && sums[(s ^ a[i])] != 0 &&
                 p[(s ^ a[i])] > a[i])
        p[(s ^ a[i])] = a[i];
  }
  int nr = 0;
  for (int s = 0; s <= 512; s++) nr += sums[s];
  cout << nr << '\n';
  for (int s = 0; s <= 512; s++)
    if (sums[s]) cout << s << ' ';
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
int T;
int N, a[100005];
bool sums[600];
int p[600];
int main() {
  fast();
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> a[i];
  sums[0] = true;
  for (int i = 1; i <= N; i++) {
    for (int s = 0; s <= 512; s++)
      if (sums[s] != 0 && p[s] < a[i] && sums[(s ^ a[i])] == 0) {
        sums[(s ^ a[i])] = true;
        p[(s ^ a[i])] = a[i];
      } else if (sums[s] != 0 && p[s] < a[i] && sums[(s ^ a[i])] != 0 &&
                 p[(s ^ a[i])] > a[i])
        p[(s ^ a[i])] = a[i];
  }
  int nr = 0;
  for (int s = 0; s <= 512; s++) nr += sums[s];
  cout << nr << '\n';
  for (int s = 0; s <= 512; s++)
    if (sums[s]) cout << s << ' ';
  return 0;
}

```