#include <bits/stdc++.h>
const long long int modd = 65536;
const long long int base = 233;
const double pi = acos(-1);
const int N = 2e5 + 10;
const int M = 1e9;
const double esp = 1e-6;
using namespace std;
long long int id = 0, n, m, tot = 0, ttt = 0, mod, a[100010], b[100010],
              c[100010];
string s;
vector<long long int> ans;
void innt() {
  memset(b, 0, sizeof(b));
  memset(c, 0, sizeof(c));
}
int main() {
  long long int zt, i, j, flat = 0, iz = 0, ip = 0, l, r, ma = 0, mi, x, y, k;
  ios::sync_with_stdio(false);
  cin >> n;
  innt();
  ans.clear();
  flat = 0;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 0) flat = 1;
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j < 520; j++) {
      c[j] = b[j];
    }
    for (j = 1; j < 520; j++) {
      if (c[j] && c[j] < a[i]) {
        x = a[i] ^ j;
        if (c[x] != 0)
          b[x] = min(a[i], c[x]);
        else
          b[x] = a[i];
      }
    }
    if (c[a[i]]) {
      flat = 1;
      b[a[i]] = min(c[a[i]], a[i]);
    } else
      b[a[i]] = a[i];
  }
  ans.push_back(0);
  for (i = 1; i < 520; i++) {
    if (b[i]) ans.push_back(i);
  }
  cout << ans.size() << endl;
  sort(ans.begin(), ans.end());
  for (i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
const long long int modd = 65536;
const long long int base = 233;
const double pi = acos(-1);
const int N = 2e5 + 10;
const int M = 1e9;
const double esp = 1e-6;
using namespace std;
long long int id = 0, n, m, tot = 0, ttt = 0, mod, a[100010], b[100010],
              c[100010];
string s;
vector<long long int> ans;
void innt() {
  memset(b, 0, sizeof(b));
  memset(c, 0, sizeof(c));
}
int main() {
  long long int zt, i, j, flat = 0, iz = 0, ip = 0, l, r, ma = 0, mi, x, y, k;
  ios::sync_with_stdio(false);
  cin >> n;
  innt();
  ans.clear();
  flat = 0;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 0) flat = 1;
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j < 520; j++) {
      c[j] = b[j];
    }
    for (j = 1; j < 520; j++) {
      if (c[j] && c[j] < a[i]) {
        x = a[i] ^ j;
        if (c[x] != 0)
          b[x] = min(a[i], c[x]);
        else
          b[x] = a[i];
      }
    }
    if (c[a[i]]) {
      flat = 1;
      b[a[i]] = min(c[a[i]], a[i]);
    } else
      b[a[i]] = a[i];
  }
  ans.push_back(0);
  for (i = 1; i < 520; i++) {
    if (b[i]) ans.push_back(i);
  }
  cout << ans.size() << endl;
  sort(ans.begin(), ans.end());
  for (i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}

```