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] 
import java.io.*;
import java.util.*;

public class q5 {
    public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    public static void solve() throws Exception {
        String[] parts = br.readLine().split(" ");
        int n = Integer.parseInt(parts[0]);

        Integer[] arr = new Integer[n];

        parts = br.readLine().split(" ");
        for (int i = 0; i < n; i++) {
            arr[i] = Integer.parseInt(parts[i]);
        }

        TreeMap<Integer,Integer> map = new TreeMap<>();
        map.put(0,0);

        for(int i = 0;i < n;i++){
            HashMap<Integer,Integer> sm = new HashMap<>();
            for(int val : map.keySet()){
                if(val >= arr[i]) break;
                if(map.get(val) < arr[i]) {
                    sm.put(val ^ arr[i],Math.min(sm.getOrDefault(val ^ arr[i],Integer.MAX_VALUE),arr[i]));
                }
            }

            for(int val : sm.keySet()) {
                map.put(val,Math.min(map.getOrDefault(val,Integer.MAX_VALUE),sm.get(val)));
            }
        }
        System.out.println(map.size());
        for(int val : map.keySet()){
            System.out.print(val + " ");
        }
        System.out.println();
    }

    public static void main(String[] args) throws Exception {

        // int tests = Integer.parseInt(br.readLine());
        // for (int test = 1; test <= tests; test++) {
            solve();
        // }
    }
}
