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: ''' Interesting Pairs
  https://codeforces.com/contest/1592/problem/A
  '''
  
  import io, os, sys
  input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline  # decode().strip() if str
  output = sys.stdout.write
  
  
  A = float('inf')
  X = 513
  
  def solve(nums, N):
  	# dp[xor] = min ending val s.t. running XOR is xor
  	dp = [A]*X
  	dp[0] = -1
  
  	for num in nums:
  		for xor in range(X):
  			if dp[xor] >= num: continue
  			dp[num^xor] = min(dp[num^xor], num)
  	return [xor for xor in range(X) if dp[xor] < A]
  
  
  def main():
  	N = int(input())
  	nums = list(map(int, input().split()))
  	out = solve(nums, N)
  	output(str(len(out)) + '\n')
  	output(' '.join(map(str, out)) + '\n')
  
  if __name__ == '__main__':
  	main()
  
  