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 F{
  	static FastReader sc=null;
  	static int nax=600;
  	
  	public static void main(String[] args) {
  		sc=new FastReader();
  		int n=sc.nextInt();
  		int a[]=sc.readArray(n);
  		
  		int dp[][]=new int[n+1][nax];
  		
  		for(int i=0;i<=n;i++)
  			Arrays.fill(dp[i], nax);
  		
  		
  		dp[0][0]=-1;
  		for(int i=0;i<n;i++) {
  			for(int j=0;j<nax;j++) {
  				dp[i+1][j]=Math.min(dp[i+1][j], dp[i][j]);
  				if(dp[i][j]<a[i]) {
  					dp[i+1][j^a[i]]=Math.min(dp[i+1][j^a[i]], a[i]);
  				}
  			}
  		}
  		
  		ArrayList<Integer> ans=new ArrayList<>();
  		
  		for(int i=0;i<nax;i++)if(dp[n][i]<nax)ans.add(i);
  		
  		System.out.println(ans.size());
  		for(int e:ans)System.out.print(e+" ");
  		System.out.println();
  		
  	}
  	
  	static int[] ruffleSort(int a[]) {
  		ArrayList<Integer> al=new ArrayList<>();
  		for(int i:a)al.add(i);
  		Collections.sort(al);
  		for(int i=0;i<a.length;i++)a[i]=al.get(i);
  		return a;
  	}
  	
  	static void print(int a[]) {
  		for(int e:a) {
  			System.out.print(e+" ");
  		}
  		System.out.println();
  	}
  	
  	static class FastReader{
  		
  		StringTokenizer st=new StringTokenizer("");
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		
  		String next() {
  			while(!st.hasMoreTokens()) 
  				try {
  					st=new StringTokenizer(br.readLine());
  				}
  			   catch(IOException e){
  				   e.printStackTrace();
  			   }
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		
  		int[] readArray(int n) {
  			int a[]=new int[n];
  			for(int i=0;i<n;i++)a[i]=sc.nextInt();
  			return a;
  		}
  	}
  	
  	
  }
  