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 sys
  import math
  import random
  import heapq
  #sys.setrecursionlimit(10**6)
  input=sys.stdin.readline
  def pow(a,n):
      if n==0:
          return 1 
      x=pow(a,n//2)
      if n%2==0 :
          return (x*x)
      else:
          return (x*x*a)
   
  def gcd(a,b):
      while b!=0:
          a,b=b,a%b
      return a 
      
  def bs(arr,le,ri):
      l=0
      arr.sort()
      r=len(arr)-1
      ans=10000000
      while l<=r:
          m=(l+r)//2
          if arr[m]>=le and arr[m]<=ri:
              ans=arr[m]
              r=m-1
          elif arr[m]<le:
              l=m+1 
          else:
              r=m-1
      return ans
   
  # t=int(input())
  t=1
  while t>0:
      n=int(input())
      a=list(map(int,input().split()))
      d={} 
      l=[512]*512
      l[0]=0
      for i in a:
          for j in range(512):
              if l[j]<i:
                  l[i^j]=min(l[i^j],i)
      ls=[]
      c=0
      for i in range(512):
          if l[i]<512:
              c+=1
              ls.append(i)
      print(c)
      print(" ".join(map(str,ls)))
      t-=1