#include <bits/stdc++.h>
using namespace std;
char gc() {
  static char buf[1 << 16], *s, *t;
  if (s == t) {
    t = (s = buf) + fread(buf, 1, 1 << 16, stdin);
    if (s == t) return EOF;
  }
  return *s++;
}
int read() {
  char c;
  int w = 1;
  while ((c = getchar()) > '9' || c < '0')
    if (c == '-') w = -1;
  int ans = c - '0';
  while ((c = getchar()) >= '0' && c <= '9')
    ans = (ans << 1) + (ans << 3) + c - '0';
  return ans * w;
}
const int xx = 1e5 + 5;
int a[xx];
vector<int> p[558];
int n;
int s[558];
int main() {
  for (int i = 0; i <= 555; i++) s[i] = 1e9;
  s[0] = 0;
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    p[a[i]].push_back(i);
  }
  for (int i = 1; i <= 555; i++) {
    for (int j = 0; j <= 555; j++) {
      if (s[j] != 1e9) {
        vector<int>::iterator it = lower_bound(p[i].begin(), p[i].end(), s[j]);
        if (it != p[i].end()) {
          s[j ^ i] = min(s[j ^ i], *it);
        }
      }
    }
  }
  int ss = 0;
  for (int i = 0; i <= 555; i++) {
    if (s[i] != 1e9) ss++;
  }
  cout << ss << "\n";
  for (int i = 0; i <= 555; i++) {
    if (s[i] != 1e9) cout << i << " ";
  }
  puts("");
  return 0;
}

### Prompt

Please create a solution in CPP 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;
char gc() {
  static char buf[1 << 16], *s, *t;
  if (s == t) {
    t = (s = buf) + fread(buf, 1, 1 << 16, stdin);
    if (s == t) return EOF;
  }
  return *s++;
}
int read() {
  char c;
  int w = 1;
  while ((c = getchar()) > '9' || c < '0')
    if (c == '-') w = -1;
  int ans = c - '0';
  while ((c = getchar()) >= '0' && c <= '9')
    ans = (ans << 1) + (ans << 3) + c - '0';
  return ans * w;
}
const int xx = 1e5 + 5;
int a[xx];
vector<int> p[558];
int n;
int s[558];
int main() {
  for (int i = 0; i <= 555; i++) s[i] = 1e9;
  s[0] = 0;
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    p[a[i]].push_back(i);
  }
  for (int i = 1; i <= 555; i++) {
    for (int j = 0; j <= 555; j++) {
      if (s[j] != 1e9) {
        vector<int>::iterator it = lower_bound(p[i].begin(), p[i].end(), s[j]);
        if (it != p[i].end()) {
          s[j ^ i] = min(s[j ^ i], *it);
        }
      }
    }
  }
  int ss = 0;
  for (int i = 0; i <= 555; i++) {
    if (s[i] != 1e9) ss++;
  }
  cout << ss << "\n";
  for (int i = 0; i <= 555; i++) {
    if (s[i] != 1e9) cout << i << " ";
  }
  puts("");
  return 0;
}

```