#include <bits/stdc++.h>
using namespace std;
using ui = unsigned int;
const double pi = acos(-1.0);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int limit = 5e5;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(long long x, char c = '\n') {
  write(x);
  putchar(c);
}
const int mod = 998244353;
const int M = 20 + 5;
const int S = (1 << 20) + 5;
const int N = 1e5 + 5;
int n, m, k, _;
int a[N];
int dp[600];
signed main() {
  while (~scanf("%d", &n)) {
    for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
    memset(dp, 0x7f7f7f7f, sizeof(dp));
    dp[0] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < 512; j++) {
        if (a[i] > dp[j]) dp[j ^ a[i]] = min(dp[j ^ a[i]], a[i]);
      }
    }
    vector<int> ans;
    for (int i = 0; i < 512; i++) {
      if (dp[i] != 0x7f7f7f7f) ans.push_back(i);
    }
    printf("%d\n", (ans.size()));
    for (auto it : ans) printf("%d ", it);
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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] 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ui = unsigned int;
const double pi = acos(-1.0);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int limit = 5e5;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(long long x, char c = '\n') {
  write(x);
  putchar(c);
}
const int mod = 998244353;
const int M = 20 + 5;
const int S = (1 << 20) + 5;
const int N = 1e5 + 5;
int n, m, k, _;
int a[N];
int dp[600];
signed main() {
  while (~scanf("%d", &n)) {
    for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
    memset(dp, 0x7f7f7f7f, sizeof(dp));
    dp[0] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < 512; j++) {
        if (a[i] > dp[j]) dp[j ^ a[i]] = min(dp[j ^ a[i]], a[i]);
      }
    }
    vector<int> ans;
    for (int i = 0; i < 512; i++) {
      if (dp[i] != 0x7f7f7f7f) ans.push_back(i);
    }
    printf("%d\n", (ans.size()));
    for (auto it : ans) printf("%d ", it);
  }
  return 0;
}

```