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 collections
import sys
import io, os
import math
from heapq import *

gcd = math.gcd
sqrt = math.sqrt


def ceil(a, b):
    a = -a
    k = a // b
    k = -k
    return k


# arr=list(map(int, input().split()))
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline


def strinp(testcases):
    k = 5
    if (testcases == -1 or testcases == 1):
        k = 1
    f = str(input())
    f = f[2:len(f) - k]
    return f


def main():
    n=int(input())
    arr=list(map(int, input().split()))
    ma=max(arr)
    came=[-1]*(512)
    came[0]=0
    for i in range(n):
        no=arr[i]
        if(came[no]==-1):
            came[no]=no
        else:
            came[no]=min(no,came[no])
        for j in range(512):
            if(came[j]!=-1 and no>came[j]):
                b=j^no
                if(came[b]==-1):
                    came[b]=no
                else:
                    came[b]=min(came[b],no)
    ans=[]
    s=0
    for i in range(512):
        if(came[i]!=-1):
            s+=1
            ans.append(i)
    print(s)
    print(*ans)
                
                    
main()