#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long INF = 1e18 + 18;
const long long MOD = 1000000007;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &s, const pair<T1, T2> &p) {
  return s << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
istream &operator>>(istream &i, vector<T> &v) {
  for (long long j = (0); j < (v.size()); j++) i >> v[j];
  return i;
}
template <typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
  int len = v.size();
  for (int i = 0; i < len; ++i) {
    s << v[i];
    if (i < len - 1) s << "	";
  }
  return s;
}
template <typename T>
ostream &operator<<(ostream &s, const vector<vector<T>> &vv) {
  s << '\n';
  int len = vv.size();
  for (int i = 0; i < len; ++i) {
    s << vv[i] << '\n';
  }
  return s;
}
int solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  long long mx = 701;
  vector<long long> dp(mx + 1, INF);
  dp[0] = 0;
  for (long long i = (0); i < (n); i++) {
    vector<long long> pre(mx + 1, INF);
    swap(pre, dp);
    for (long long j = (0); j < (mx + 1); j++) {
      long long v = a[i];
      if (pre[j] < v) {
        if ((j ^ v) <= 701) {
          chmin(dp[j ^ v], v);
        }
      }
      chmin(dp[j], pre[j]);
    }
  }
  vector<long long> ans;
  for (long long i = (0); i < (mx); i++) {
    if (dp[i] != INF) {
      ans.push_back(i);
    }
  }
  cout << ans.size() << '\n';
  for (long long i = (0); i < (ans.size()); i++) {
    cout << ans[i] << " ";
  }
  cout << '\n';
  return 0;
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << std::setprecision(10);
  solve();
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long INF = 1e18 + 18;
const long long MOD = 1000000007;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &s, const pair<T1, T2> &p) {
  return s << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
istream &operator>>(istream &i, vector<T> &v) {
  for (long long j = (0); j < (v.size()); j++) i >> v[j];
  return i;
}
template <typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
  int len = v.size();
  for (int i = 0; i < len; ++i) {
    s << v[i];
    if (i < len - 1) s << "	";
  }
  return s;
}
template <typename T>
ostream &operator<<(ostream &s, const vector<vector<T>> &vv) {
  s << '\n';
  int len = vv.size();
  for (int i = 0; i < len; ++i) {
    s << vv[i] << '\n';
  }
  return s;
}
int solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  long long mx = 701;
  vector<long long> dp(mx + 1, INF);
  dp[0] = 0;
  for (long long i = (0); i < (n); i++) {
    vector<long long> pre(mx + 1, INF);
    swap(pre, dp);
    for (long long j = (0); j < (mx + 1); j++) {
      long long v = a[i];
      if (pre[j] < v) {
        if ((j ^ v) <= 701) {
          chmin(dp[j ^ v], v);
        }
      }
      chmin(dp[j], pre[j]);
    }
  }
  vector<long long> ans;
  for (long long i = (0); i < (mx); i++) {
    if (dp[i] != INF) {
      ans.push_back(i);
    }
  }
  cout << ans.size() << '\n';
  for (long long i = (0); i < (ans.size()); i++) {
    cout << ans[i] << " ";
  }
  cout << '\n';
  return 0;
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << std::setprecision(10);
  solve();
}

```