#include <bits/stdc++.h>
using namespace std;
const int siz1 = 1e5, siz2 = 512, siz3 = 1e6, mod = 998244353;
int n, minidx[siz2 + 5];
pair<int, int> arr[siz1 + 5];
bool ans[siz2 + 5];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &arr[i].first);
    arr[i].second = i;
  }
  for (int i = 0; i < siz2; ++i) {
    minidx[i] = INT_MAX;
  }
  sort(arr, arr + n);
  for (int i = 0; i < n; ++i) {
    ans[arr[i].first] = 1;
    minidx[arr[i].first] = min(minidx[arr[i].first], arr[i].second);
    for (int j = 0; j < siz2; ++j) {
      if (minidx[j] < arr[i].second) {
        int val = j ^ arr[i].first;
        ans[val] = 1;
        minidx[val] = min(minidx[val], arr[i].second);
      }
    }
  }
  ans[0] = 1;
  int ans_sz = 0;
  for (int i = 0; i < siz2; ++i) {
    if (ans[i]) {
      ++ans_sz;
    }
  }
  printf("%d\n", ans_sz);
  for (int i = 0; i < siz2; ++i) {
    if (ans[i]) {
      printf("%d ", i);
    }
  }
  printf("\n");
  return 0;
}

### Prompt

Please create a solution in cpp 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 siz1 = 1e5, siz2 = 512, siz3 = 1e6, mod = 998244353;
int n, minidx[siz2 + 5];
pair<int, int> arr[siz1 + 5];
bool ans[siz2 + 5];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &arr[i].first);
    arr[i].second = i;
  }
  for (int i = 0; i < siz2; ++i) {
    minidx[i] = INT_MAX;
  }
  sort(arr, arr + n);
  for (int i = 0; i < n; ++i) {
    ans[arr[i].first] = 1;
    minidx[arr[i].first] = min(minidx[arr[i].first], arr[i].second);
    for (int j = 0; j < siz2; ++j) {
      if (minidx[j] < arr[i].second) {
        int val = j ^ arr[i].first;
        ans[val] = 1;
        minidx[val] = min(minidx[val], arr[i].second);
      }
    }
  }
  ans[0] = 1;
  int ans_sz = 0;
  for (int i = 0; i < siz2; ++i) {
    if (ans[i]) {
      ++ans_sz;
    }
  }
  printf("%d\n", ans_sz);
  for (int i = 0; i < siz2; ++i) {
    if (ans[i]) {
      printf("%d ", i);
    }
  }
  printf("\n");
  return 0;
}

```