#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    long long sum = 0;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    vector<long long> st(11, 1);
    for (int i = 1; i < 11; ++i) {
      st[i] = st[i - 1] * 2;
    }
    vector<pair<long long, long long>> dp(513, {0, 550});
    dp[0].first = 1;
    dp[0].second = 513;
    for (int i = 0; i < n; ++i) {
      long long k;
      for (int j = 0; j < 11; ++j) {
        if (st[j] >= a[i]) {
          k = st[j];
          break;
        }
      }
      if (dp[a[i]].first) {
        dp[a[i]].second = min(dp[a[i]].second, a[i]);
      } else {
        dp[a[i]].first = 1;
        dp[a[i]].second = a[i];
      }
      for (int j = 0; j <= k; ++j) {
        if (dp[j].second < a[i]) {
          if (dp[j ^ a[i]].first == 0) {
            dp[j ^ a[i]].first = 1;
            dp[j ^ a[i]].second = a[i];
          } else {
            dp[j ^ a[i]].second = min(dp[j ^ a[i]].second, a[i]);
          }
        }
      }
    }
    long long d = 0;
    for (int i = 0; i < 513; ++i) {
      if (dp[i].first) {
        ++d;
      }
    }
    cout << d << '\n';
    for (int i = 0; i < 513; ++i) {
      if (dp[i].first) {
        cout << i << " ";
      }
    }
  }
}

### Prompt

Your challenge is to write a cpp solution 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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    long long sum = 0;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    vector<long long> st(11, 1);
    for (int i = 1; i < 11; ++i) {
      st[i] = st[i - 1] * 2;
    }
    vector<pair<long long, long long>> dp(513, {0, 550});
    dp[0].first = 1;
    dp[0].second = 513;
    for (int i = 0; i < n; ++i) {
      long long k;
      for (int j = 0; j < 11; ++j) {
        if (st[j] >= a[i]) {
          k = st[j];
          break;
        }
      }
      if (dp[a[i]].first) {
        dp[a[i]].second = min(dp[a[i]].second, a[i]);
      } else {
        dp[a[i]].first = 1;
        dp[a[i]].second = a[i];
      }
      for (int j = 0; j <= k; ++j) {
        if (dp[j].second < a[i]) {
          if (dp[j ^ a[i]].first == 0) {
            dp[j ^ a[i]].first = 1;
            dp[j ^ a[i]].second = a[i];
          } else {
            dp[j ^ a[i]].second = min(dp[j ^ a[i]].second, a[i]);
          }
        }
      }
    }
    long long d = 0;
    for (int i = 0; i < 513; ++i) {
      if (dp[i].first) {
        ++d;
      }
    }
    cout << d << '\n';
    for (int i = 0; i < 513; ++i) {
      if (dp[i].first) {
        cout << i << " ";
      }
    }
  }
}

```