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.util.*;

public class HelloWorld {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = 1;
        StringBuilder ret = new StringBuilder();
        while(t-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for(int i = 0; i < n; i++) a[i] = sc.nextInt();
            int[] dp = new int[514];
            Arrays.fill(dp, 600);
            dp[0] = 0;
            for(int i = 0; i < n; i++) {
                for(int j = 0; j <= 512; j++) {
                    if(dp[j] < a[i]) {
                        dp[j ^ a[i]] = Math.min(dp[j ^ a[i]], a[i]);
                    }
                }
            }
            ArrayList<Integer> arr = new ArrayList<>();
            for(int i = 0; i <= 512; i++) {
                if(dp[i] < 600) arr.add(i);
            }
            ret.append(arr.size() + "\n");
            for(int aa:arr) ret.append(aa + " ");
            ret.append("\n");
        }
        System.out.println(ret);
    }
}