#include <bits/stdc++.h>
using namespace std;
const int M = 9;
const int N = 100100;
void minimize(int& x, int y) {
  if (x > y) {
    x = y;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int& x : a) {
    scanf("%d", &x);
  }
  vector<int> b(n);
  iota(b.begin(), b.end(), 0);
  sort(b.begin(), b.end(), [&a](int i, int j) {
    return make_pair(a[i], -i) < make_pair(a[j], -j);
  });
  vector<int> minimum(1 << M, n);
  minimum[0] = -1;
  for (int i : b) {
    for (int j = 0; j < (1 << M); j++) {
      if (minimum[j] < i) {
        minimize(minimum[a[i] ^ j], i);
      }
    }
  }
  vector<int> answer;
  for (int j = 0; j < (1 << M); j++) {
    if (minimum[j] < n) {
      answer.push_back(j);
    }
  }
  cout << answer.size() << endl;
  for (int j : answer) {
    printf("%d ", j);
  }
  puts("");
}

### 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 M = 9;
const int N = 100100;
void minimize(int& x, int y) {
  if (x > y) {
    x = y;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int& x : a) {
    scanf("%d", &x);
  }
  vector<int> b(n);
  iota(b.begin(), b.end(), 0);
  sort(b.begin(), b.end(), [&a](int i, int j) {
    return make_pair(a[i], -i) < make_pair(a[j], -j);
  });
  vector<int> minimum(1 << M, n);
  minimum[0] = -1;
  for (int i : b) {
    for (int j = 0; j < (1 << M); j++) {
      if (minimum[j] < i) {
        minimize(minimum[a[i] ^ j], i);
      }
    }
  }
  vector<int> answer;
  for (int j = 0; j < (1 << M); j++) {
    if (minimum[j] < n) {
      answer.push_back(j);
    }
  }
  cout << answer.size() << endl;
  for (int j : answer) {
    printf("%d ", j);
  }
  puts("");
}

```