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] 
#include <bits/stdc++.h>
using namespace std;
int dx[] = {-1, 0, 1, 0, 0, 0, 0, 0};
int dy[] = {0, 1, 0, -1, 0, 0, 0, 0};
long long n;
vector<long long> a;
long long can[512][512];
long long dp[550][550];
long long recur(long long node, long long cur) {
  if (cur == 0) return 1;
  if (node == 511) {
    return 0;
  }
  if (dp[node][cur] != -1) return dp[node][cur];
  long long ans = 0;
  for (int i = node + 1; i < 512; i++) {
    if (can[node][i] == 1) {
      ans = ans | recur(i, cur ^ i);
    }
  }
  return dp[node][cur] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  a.push_back(0);
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a.push_back(x);
  }
  n = a.size();
  long long f[512];
  memset(f, 0, sizeof(f));
  memset(can, 0, sizeof(can));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < a[i]; j++) {
      if (f[j] == 1) {
        can[j][a[i]] = 1;
      }
    }
    f[a[i]] = 1;
  }
  memset(dp, -1, sizeof(dp));
  vector<long long> ans;
  for (int i = 0; i < 512; i++) {
    if (recur(0, i)) ans.push_back(i);
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
