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] 
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