#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 11;
const long long M = 1e6 + 21;
const long long big = 1e17;
const long long hsh2 = 1964325029;
const long long mod = 1e9 + 7;
const double EPS = 1e-9;
const long long block = 400;
const long long shift = 2e3;
const double pi = acos(-1.0);
mt19937_64 bruh(chrono::steady_clock::now().time_since_epoch().count());
long long n;
long long a[N];
long long dp[512];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i <= 511; i++) dp[i] = 501;
  dp[0] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    for (int j = 0; j < 512; j++) {
      if (a[i] > dp[j]) dp[j ^ a[i]] = min(dp[j ^ a[i]], a[i]);
    }
  }
  long long cnt = 0;
  for (int i = 0; i <= 511; i++) cnt += (dp[i] != 501);
  cout << cnt << '\n';
  for (int i = 0; i <= 511; i++)
    if (dp[i] != 501) cout << i << " ";
  cout << '\n';
}

### Prompt

Create a solution in cpp for 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 long long N = 2e5 + 11;
const long long M = 1e6 + 21;
const long long big = 1e17;
const long long hsh2 = 1964325029;
const long long mod = 1e9 + 7;
const double EPS = 1e-9;
const long long block = 400;
const long long shift = 2e3;
const double pi = acos(-1.0);
mt19937_64 bruh(chrono::steady_clock::now().time_since_epoch().count());
long long n;
long long a[N];
long long dp[512];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i <= 511; i++) dp[i] = 501;
  dp[0] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    for (int j = 0; j < 512; j++) {
      if (a[i] > dp[j]) dp[j ^ a[i]] = min(dp[j ^ a[i]], a[i]);
    }
  }
  long long cnt = 0;
  for (int i = 0; i <= 511; i++) cnt += (dp[i] != 501);
  cout << cnt << '\n';
  for (int i = 0; i <= 511; i++)
    if (dp[i] != 501) cout << i << " ";
  cout << '\n';
}

```