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.Arrays;
import java.util.Scanner;

public class E1582F1 {
    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);
        int n = scn.nextInt();

        int[] lastVal = new int[501];
        Arrays.fill(lastVal, 1000);
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = scn.nextInt();
            lastVal[arr[i]] = Math.min(arr[i], lastVal[arr[i]]);
            for (int j = 0; j < 501 && j < arr[i]; j++) {
                if (lastVal[j] < arr[i]) {
                    lastVal[arr[i] ^ j] = Math.min(arr[i], lastVal[arr[i] ^ j]);
                }
            }
        }
        lastVal[0] = 1;
        StringBuilder sb = new StringBuilder();
        int count = 0;
        for (int v : lastVal) {
            if (v < 1000) count++;
        }
        sb.append(count);
        sb.append("\n");
        for (int i = 0; i < lastVal.length; i++) {
            if (lastVal[i] < 1000) {
                sb.append(i);
                sb.append(" ");
            }
        }
        System.out.println(sb);

    }
}
