description: 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] 
solution: import java.io.*;
  import java.util.*;
  
  public class new1{
  	public static int func(int[][] dp, int[] arr, int x, int y) {
  		
  		int n = arr.length;
  		if(x == n) {
  			//System.out.println(x + " " + y );
  			if(y == 0) return 501;
  			else return -1;
  		}
  		if(dp[x][y] != -2) return dp[x][y];
  		int a1 = -1;
  		int aa1 = func(dp, arr, x + 1, y ^ arr[x]);
  		if(aa1 != -1 && aa1 > arr[x]) a1 = arr[x];
  		int a2 = func(dp, arr, x + 1, y);
  		int ans = -1;
  		if(a1 == -1 && a2 != -1) ans = a2;
  		else if(a2 == -1 && a1 != -1) ans = a1;
  		else if(a1 != - 1&& a2 != -1) ans = Math.max(a1, a2);
  		dp[x][y] = ans;
  		//System.out.println(x + " " + y + " " + ans);
  		return ans;
  		
  	}
       
  
  	public static void main(String[] args)  throws IOException{
      	FastReader s = new FastReader();
      	BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
      	int t =  1;//s.nextInt();
      	int bit = 512;
      	//long l1 = System.currentTimeMillis();
      	for(int z = 0; z < t; z++) {
      		int n = s.nextInt();
      		int[] arr = new int[n];
      		for(int i = 0; i < n; i++) {
      			arr[i] = s.nextInt();
      		}
      		int[][] dp = new int[n][bit];
      		for(int i = 0; i < n; i++) {
      			Arrays.fill(dp[i], -2);
      		}
      		int count = 0;
      		int[] ans = new int[bit]; 
      		for(int i = n; i >= 0; i--) {
  	    		for(int j = 0; j < bit; j++) {
  	    			if(func(dp, arr, i, j) != -1) {
  	    				if(ans[j] != 1) count++;
  	    				ans[j] = 1;
  	    			}
  	    		}
      		}
  //    		for(int i = 0; i < n; i++) {
  //    			System.out.println(Arrays.toString(dp[i]));
  //    		}
      		System.out.println(count + " ");
      		for(int i = 0; i < bit; i++) {
      			if(ans[i] == 1) System.out.print(i + " "); 
      		}
      		System.out.println();
      	}
      	//System.out.println(System.currentTimeMillis() - l1);
      			//output.flush();	
      }
  }
  
  class FastReader {
      BufferedReader br;
      StringTokenizer st;
      public FastReader() {
          br = new BufferedReader(new InputStreamReader(System.in));
      }
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
   
      public int nextInt() {
          return Integer.parseInt(next());
      }
   
      long nextLong() {
          return Long.parseLong(next());
      }
   
      double nextDouble() {
          return Double.parseDouble(next());
      }
   
      String nextLine() {
          String str = "";
          try {
              str = br.readLine();
          } catch (IOException e) {
              e.printStackTrace();
          }
          return str;
  }}