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 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();
	}
}
