#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0) {
    return a + b;
  }
  return gcd(b, a % b);
}
int32_t main() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  long long MX = 1 << 10;
  vector<long long> ans(MX, -1);
  ans[0] = 0;
  long long tmp;
  long long count = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < MX; j++) {
      if (ans[j] != -1 && v[i] > ans[j]) {
        tmp = j ^ v[i];
        if (ans[tmp] == -1) {
          ans[tmp] = v[i];
          count++;
        } else {
          ans[tmp] = min(ans[tmp], v[i]);
        }
      }
    }
    if (ans[v[i]] == -1) {
      count++;
      ans[v[i]] = v[i];
    } else {
      ans[v[i]] = min(ans[v[i]], v[i]);
    }
  }
  count++;
  cout << count << '\n';
  for (long long i = 0; i < MX; i++) {
    if (ans[i] != -1) {
      cout << i << " ";
    }
  }
  cout << '\n';
}

### 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;
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0) {
    return a + b;
  }
  return gcd(b, a % b);
}
int32_t main() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  long long MX = 1 << 10;
  vector<long long> ans(MX, -1);
  ans[0] = 0;
  long long tmp;
  long long count = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < MX; j++) {
      if (ans[j] != -1 && v[i] > ans[j]) {
        tmp = j ^ v[i];
        if (ans[tmp] == -1) {
          ans[tmp] = v[i];
          count++;
        } else {
          ans[tmp] = min(ans[tmp], v[i]);
        }
      }
    }
    if (ans[v[i]] == -1) {
      count++;
      ans[v[i]] = v[i];
    } else {
      ans[v[i]] = min(ans[v[i]], v[i]);
    }
  }
  count++;
  cout << count << '\n';
  for (long long i = 0; i < MX; i++) {
    if (ans[i] != -1) {
      cout << i << " ";
    }
  }
  cout << '\n';
}

```