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 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)