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: from collections import defaultdict
  def il(): #Input list
      return list(map(int,input().split()))
  def ii(): #Input int
      return int(input())
  def imi(): #Input multiple int
      return map(int,input().split())
  def ist(): #Input String
      return input()
  
  n=ii()
  ar=il()
  
  li=defaultdict(lambda :0)
  
  li[0]=1
  g=[[0] for i in range(501)]
  pg=[defaultdict(lambda :0) for j in range(501)]
  pr=defaultdict(lambda :0)
  
  for i in ar:
      pr[i]=1
      pg[i][0]=1
  
  
  for a in ar:
      for f in g[a]:
          x=a^f
          li[x]=1
          for b in range(a+1,501):
              if pr[b]:
                  if pg[b][x]==0:
                      pg[b][x]=1
                      g[b].append(x)
                  else: break
      g[a]=[]
  
  la=[]
  l=0
  for i in range(250007):
      if li[i]:
          la.append(i)
          l+=1
  print(l)
  print(*la)
  
  
  