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] 
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define int long long
#define rep(i, a, b) for(int i=a; i<b; i++)
#define rrep(i, a, b) for(int i=a; i>=b; i--)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

typedef tree<int, null_type, greater_equal<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;

void solve(){

    int n;

    cin >> n;

    vector<int> a(n);
    set<int> st;
    st.insert(0);

    rep(i, 0, n){
        cin >> a[i];
    }

    rep(i, 0, n){
        rep(j, 0, 505){
            if(st.find(j) != st.end()) continue;
            int num = a[i] ^ j;
            if(num < a[i] && st.find(num) != st.end()){
                st.insert(j);
            }
        }
        st.insert(a[i]);
    }

    cout << st.size() << endl;
    for(auto x : st) cout << x << " ";
    cout << endl;
}

int32_t main(){

    IOS;

    int t;

	t = 1;

    rep(i, 1, t+1) solve();
}