#include <bits/stdc++.h>
int const maxn = 100003;
int const V = 1 << 9, inf = 0x3f3f3f3f;
int n = 0;
int a[maxn];
bool f[V];
int minPos[V];
inline int min(int a, int b) { return a < b ? a : b; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 0; i < V; ++i) minPos[i] = inf;
  minPos[0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < V; ++j) f[j] = (minPos[j ^ a[i]] < a[i]);
    for (int j = 0; j < V; ++j)
      if (f[j]) minPos[j] = min(minPos[j], a[i]);
  }
  int k = 0;
  for (int i = 0; i < V; ++i)
    if (minPos[i] != inf) ++k;
  printf("%d\n", k);
  for (int i = 0; i < V; ++i)
    if (minPos[i] != inf) printf("%d ", i);
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
int const maxn = 100003;
int const V = 1 << 9, inf = 0x3f3f3f3f;
int n = 0;
int a[maxn];
bool f[V];
int minPos[V];
inline int min(int a, int b) { return a < b ? a : b; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 0; i < V; ++i) minPos[i] = inf;
  minPos[0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < V; ++j) f[j] = (minPos[j ^ a[i]] < a[i]);
    for (int j = 0; j < V; ++j)
      if (f[j]) minPos[j] = min(minPos[j], a[i]);
  }
  int k = 0;
  for (int i = 0; i < V; ++i)
    if (minPos[i] != inf) ++k;
  printf("%d\n", k);
  for (int i = 0; i < V; ++i)
    if (minPos[i] != inf) printf("%d ", i);
  return 0;
}

```