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 Main {
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static StringTokenizer st;
  	static PrintWriter pr = new PrintWriter(System.out);
  	static String readLine() throws IOException {
  		return br.readLine();
  	}
  	static String next() throws IOException {
  		while (st == null || !st.hasMoreTokens())
  			st = new StringTokenizer(readLine());
  		return st.nextToken();
  	}
  	static int readInt() throws IOException {
  		return Integer.parseInt(next());
  	}
  	static long readLong() throws IOException {
  		return Long.parseLong(next());
  	}
  	static double readDouble() throws IOException {
  		return Double.parseDouble(next());
  	}
  	static char readChar() throws IOException {
  		return next().charAt(0);
  	}
  	static class Pair implements Comparable<Pair> {
  		int f, s;
  		Pair(int f, int s) {
  			this.f = f; this.s = s;
  		}
  		public int compareTo(Pair other) {
  			if (this.f != other.f) return this.f - other.f;
  			return this.s - other.s;
  		}
  	}
  	static void solve() throws IOException {
  		int n = readInt();
  		boolean dp[][] = new boolean[512][512];
  		//[possible value of xor][value of element]
  		Arrays.fill(dp[0], true);
  		for (int i = 1; i <= n; ++i) {
  			int a = readInt();
  			if (a == 0) continue;
  			for (int x = 0; x < 512; ++x) {
  				if (dp[x][a - 1]) {
  					int y = x^a;
  					for (int k = a; k < 512 && !dp[y][k]; ++k) dp[y][k] = true;
  				}
  			}
  		}
  		int ans = 0;
  		for (int i = 0; i < 512; ++i)
  			if (dp[i][511]) ++ans;
  		pr.println(ans);
  		for (int i = 0; i < 512; ++i)
  			if (dp[i][511]) pr.print(i + " ");
  		pr.println();
  	}
  	public static void main(String[] args) throws IOException {
  		solve();
  		//for (int t = readInt(); t > 0; --t) solve();
  		pr.close();
  	}
  }
  