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 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()