#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const int MOD = 1e9 + 7;
int binarySearch(int arr[], int l, int r, int x) {
  if (r >= l) {
    int mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  } else {
    return -1;
  }
}
ll power(ll x, ll y) {
  ll res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
const int N = 100000;
int arr[N];
int mn[512];
void solve() {
  int n;
  cin >> n;
  for (int i = (0); i < (n); i++) {
    cin >> arr[i];
  }
  mn[0] = -1;
  for (int i = (0); i < (n); i++) {
    for (int j = (0); j < (512); j++) {
      if (mn[j] != MOD && arr[i] > mn[j]) {
        int y = arr[i] ^ j;
        mn[y] = min(mn[y], arr[i]);
      }
    }
  }
  int ans = 0;
  vector<int> val;
  for (int i = (0); i < (512); i++) {
    if (mn[i] != MOD) {
      ans++;
      val.push_back(i);
    }
  }
  cout << ans << '\n';
  for (int i = (0); i < (ans); i++) {
    cout << val[i] << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int i = (0); i < (512); i++) {
    mn[i] = MOD;
  }
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
using ll = long long;
using ld = long double;
const int MOD = 1e9 + 7;
int binarySearch(int arr[], int l, int r, int x) {
  if (r >= l) {
    int mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  } else {
    return -1;
  }
}
ll power(ll x, ll y) {
  ll res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
const int N = 100000;
int arr[N];
int mn[512];
void solve() {
  int n;
  cin >> n;
  for (int i = (0); i < (n); i++) {
    cin >> arr[i];
  }
  mn[0] = -1;
  for (int i = (0); i < (n); i++) {
    for (int j = (0); j < (512); j++) {
      if (mn[j] != MOD && arr[i] > mn[j]) {
        int y = arr[i] ^ j;
        mn[y] = min(mn[y], arr[i]);
      }
    }
  }
  int ans = 0;
  vector<int> val;
  for (int i = (0); i < (512); i++) {
    if (mn[i] != MOD) {
      ans++;
      val.push_back(i);
    }
  }
  cout << ans << '\n';
  for (int i = (0); i < (ans); i++) {
    cout << val[i] << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int i = (0); i < (512); i++) {
    mn[i] = MOD;
  }
  while (t--) {
    solve();
  }
  return 0;
}

```