#include <bits/stdc++.h>
int main() {
  int32_t num_tests;
  num_tests = 1;
  std::mt19937 rng;
  for (int32_t t = 0; t < num_tests; t++) {
    int32_t n;
    std::cin >> n;
    int32_t* arr = new int32_t[n];
    for (int32_t i = 0; i < n; i++) std::cin >> arr[i];
    int32_t* min_end_to_reach = new int32_t[512];
    for (int32_t i = 1; i < 512; i++) min_end_to_reach[i] = INT32_MAX;
    min_end_to_reach[0] = 0;
    for (int32_t i = 0; i < n; i++) {
      for (int32_t j = 0; j < 512; j++) {
        int32_t want_prev = arr[i] ^ j;
        if (min_end_to_reach[want_prev] < arr[i])
          min_end_to_reach[j] = std::min(min_end_to_reach[j], arr[i]);
      }
    }
    std::vector<int32_t> ans;
    for (int32_t i = 0; i < 512; i++)
      if (min_end_to_reach[i] != INT32_MAX) ans.push_back(i);
    std::cout << ans.size() << "\n";
    for (int32_t i = 0; i < ans.size(); i++) std::cout << ans[i] << " ";
    std::cout << "\n";
  }
  return 0;
}

### 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>
int main() {
  int32_t num_tests;
  num_tests = 1;
  std::mt19937 rng;
  for (int32_t t = 0; t < num_tests; t++) {
    int32_t n;
    std::cin >> n;
    int32_t* arr = new int32_t[n];
    for (int32_t i = 0; i < n; i++) std::cin >> arr[i];
    int32_t* min_end_to_reach = new int32_t[512];
    for (int32_t i = 1; i < 512; i++) min_end_to_reach[i] = INT32_MAX;
    min_end_to_reach[0] = 0;
    for (int32_t i = 0; i < n; i++) {
      for (int32_t j = 0; j < 512; j++) {
        int32_t want_prev = arr[i] ^ j;
        if (min_end_to_reach[want_prev] < arr[i])
          min_end_to_reach[j] = std::min(min_end_to_reach[j], arr[i]);
      }
    }
    std::vector<int32_t> ans;
    for (int32_t i = 0; i < 512; i++)
      if (min_end_to_reach[i] != INT32_MAX) ans.push_back(i);
    std::cout << ans.size() << "\n";
    for (int32_t i = 0; i < ans.size(); i++) std::cout << ans[i] << " ";
    std::cout << "\n";
  }
  return 0;
}

```