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 tank {

    static final FastScanner fs = new FastScanner();

    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        int t = 1;
        while(t-- > 0) run_case();

        out.close();
    }

    static void run_case() {
        int n = fs.nextInt();
        int[] a = fs.readArray(n);
        int[][] cnt = new int[513][2];

        cnt[0][0] = 1;
        cnt[0][1] = 0;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 513; j++) {
                if(cnt[j][0] > 0 && cnt[j][1] < a[i]) {
                    cnt[a[i] ^ j][0]++;
                    if(cnt[a[i] ^ j][0] > 1)
                        cnt[a[i] ^ j][1] = Math.min(a[i], cnt[a[i] ^ j][1]);
                    else
                        cnt[a[i] ^ j][1] = a[i];
                }
            }
        }

        int k = 0;

        for (int i = 0; i < 513; i++) {
            if(cnt[i][0] > 0) k++;
        }

        out.println(k);

        for (int i = 0; i < 513; i++) {
            if(cnt[i][0] > 0) out.print(i + " ");
        }

        out.println();
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        String nextLine(){
            try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }

            return "";
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}