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 bisect import bisect_left,bisect_right
  def findppos(arr,lm,k,p,a,i,j):
      ln=lm[k]
      x=bisect_right(arr[k],p)
      if x==ln:
          return False
      if a[k][j^k]==-1:
          a[k][j^k]=arr[k][x]
      else:
          a[k][j^k]=min(a[k][j^k],arr[k][x])
      return True
      
      
  vis=[0 for i in range(512)]
  vis[0]=1
  v=[10000000 for i in range(512)]
  n=int(input())
  arr=list(map(int,input().split()))
  a=[[-1 for i in range(512)] for j in range(512)]
  l=[[] for i in range(501)]
  lm=[0 for i in range(501)]
  for i in range(n):
      l[arr[i]].append(i)
      lm[arr[i]]+=1
      vis[arr[i]]=1
      if a[arr[i]][arr[i]]==-1:
          a[arr[i]][arr[i]]=i
  # print(lm)
  for i in range(501):
      for j in range(512):
          if a[i][j]!=-1 and a[i][j]<v[j]:
              v[j]=a[i][j]
              for k in range(i+1,501):
                  pos=findppos(l,lm,k,a[i][j],a,i,j)
                  if pos:
                      x=j^k
                      vis[x]=1
  print(sum(vis))
  # print(a[2])
  ar=[]
  for i in range(512):
      if vis[i]==1:
          ar.append(i)
  print(*ar)