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] 
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=[]
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]:
        pg[a][f]=0
        x=a^f
        li.append(x)
        for b in range(a+1,502):
            if pr[b]:
                if pg[b][x]==0:
                    pg[b][x]=1
                    g[b].append(x)
    g[a]=[]
li=sorted(list(set(li)))
print(len(li))
print(*li)


