#include <bits/stdc++.h>
using namespace std;
inline int read() {
  register int x = 0, f = 0;
  register char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - 48;
    c = getchar();
  }
  return f ? -x : x;
}
const int N = 5e3 + 3, M = 5e3 + 3, eps = 1e-4;
int n, num;
int vis[600];
signed main() {
  n = read();
  for (register int i = 1; i <= n; ++i) {
    int x = read();
    if (!vis[x])
      vis[x] = x;
    else
      vis[x] = min(x, vis[x]);
    for (register int j = 1; j <= 511; ++j) {
      if (!vis[j]) continue;
      if (x <= vis[j]) continue;
      int y = j ^ x;
      if (!vis[y])
        vis[y] = x;
      else
        vis[y] = min(vis[y], x);
    }
  }
  int ans = 1;
  for (register int i = 1; i <= 511; ++i)
    if (vis[i] > 0) ans++;
  printf("%d\n0 ", ans);
  for (register int i = 1; i <= 511; ++i)
    if (vis[i] > 0) printf("%d ", i);
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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>
using namespace std;
inline int read() {
  register int x = 0, f = 0;
  register char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - 48;
    c = getchar();
  }
  return f ? -x : x;
}
const int N = 5e3 + 3, M = 5e3 + 3, eps = 1e-4;
int n, num;
int vis[600];
signed main() {
  n = read();
  for (register int i = 1; i <= n; ++i) {
    int x = read();
    if (!vis[x])
      vis[x] = x;
    else
      vis[x] = min(x, vis[x]);
    for (register int j = 1; j <= 511; ++j) {
      if (!vis[j]) continue;
      if (x <= vis[j]) continue;
      int y = j ^ x;
      if (!vis[y])
        vis[y] = x;
      else
        vis[y] = min(vis[y], x);
    }
  }
  int ans = 1;
  for (register int i = 1; i <= 511; ++i)
    if (vis[i] > 0) ans++;
  printf("%d\n0 ", ans);
  for (register int i = 1; i <= 511; ++i)
    if (vis[i] > 0) printf("%d ", i);
  return 0;
}

```