#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N], ans[N], f[700][700];
int lowbit(int x) { return x & (-x); }
int n, c[1100][1100];
int query(int u, int v) {
  int ans = 0;
  while (u > 0) {
    ans += c[v][u];
    u -= lowbit(u);
  }
  return ans;
}
void add(int u, int v) {
  while (u <= 1025) {
    c[v][u]++;
    u += lowbit(u);
  }
}
int main() {
  add(1, 0);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    for (int j = 0; j <= 512; j++) {
      if (query(a[i], j) > 0) add(a[i] + 1, a[i] ^ j);
    }
    add(a[i] + 1, a[i]);
  }
  vector<int> ans;
  for (int i = 0; i <= 1025; i++)
    if (query(1025, i) > 0) ans.push_back(i);
  printf("%d\n", ans.size());
  for (auto x : ans) printf("%d ", x);
  puts("");
  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 int N = 1e5 + 10;
int a[N], ans[N], f[700][700];
int lowbit(int x) { return x & (-x); }
int n, c[1100][1100];
int query(int u, int v) {
  int ans = 0;
  while (u > 0) {
    ans += c[v][u];
    u -= lowbit(u);
  }
  return ans;
}
void add(int u, int v) {
  while (u <= 1025) {
    c[v][u]++;
    u += lowbit(u);
  }
}
int main() {
  add(1, 0);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    for (int j = 0; j <= 512; j++) {
      if (query(a[i], j) > 0) add(a[i] + 1, a[i] ^ j);
    }
    add(a[i] + 1, a[i]);
  }
  vector<int> ans;
  for (int i = 0; i <= 1025; i++)
    if (query(1025, i) > 0) ans.push_back(i);
  printf("%d\n", ans.size());
  for (auto x : ans) printf("%d ", x);
  puts("");
  return 0;
}

```