#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
long long int pow_mod(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return res;
}
long long int inv_mod(long long int a) { return pow_mod(a, mod - 2); }
void solve() {
  long long int n;
  cin >> n;
  long long int a[n + 1];
  for (long long int i = 0; i < n; i++) cin >> a[i + 1];
  long long int mn[512];
  for (long long int i = 0; i < 512; i++) mn[i] = (long long int)1e18;
  mn[0] = 0;
  set<long long int> st;
  for (long long int pos = 1; pos <= n; pos++) {
    for (long long int xr = 0; xr < 512; xr++) {
      long long int rem = xr ^ a[pos];
      if (mn[rem] < a[pos]) {
        st.insert(xr);
        mn[xr] = min(mn[xr], a[pos]);
      }
    }
  }
  st.insert(0);
  vector<long long int> ans;
  for (auto e : st) ans.push_back(e);
  cout << ans.size() << "\n";
  for (auto e : ans) cout << e << " ";
  cout << endl;
}
int main() {
  long long int t;
  t = 1;
  while (t--) solve();
  return 0;
}

### 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>
using namespace std;
const long long int mod = 1000000007;
long long int pow_mod(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return res;
}
long long int inv_mod(long long int a) { return pow_mod(a, mod - 2); }
void solve() {
  long long int n;
  cin >> n;
  long long int a[n + 1];
  for (long long int i = 0; i < n; i++) cin >> a[i + 1];
  long long int mn[512];
  for (long long int i = 0; i < 512; i++) mn[i] = (long long int)1e18;
  mn[0] = 0;
  set<long long int> st;
  for (long long int pos = 1; pos <= n; pos++) {
    for (long long int xr = 0; xr < 512; xr++) {
      long long int rem = xr ^ a[pos];
      if (mn[rem] < a[pos]) {
        st.insert(xr);
        mn[xr] = min(mn[xr], a[pos]);
      }
    }
  }
  st.insert(0);
  vector<long long int> ans;
  for (auto e : st) ans.push_back(e);
  cout << ans.size() << "\n";
  for (auto e : ans) cout << e << " ";
  cout << endl;
}
int main() {
  long long int t;
  t = 1;
  while (t--) solve();
  return 0;
}

```