#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, fh = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * fh;
}
inline void out(int *a, int l, int r) {
  for (int i = l; i <= r; ++i) printf("%d ", *(a + i));
  puts("");
}
const int N = 1e5 + 5, inf = 1e9;
int a[N], b[N], n, ok[N];
bool f[N][550];
signed main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 0; i <= 1024; ++i) b[i] = inf;
  ok[0] = 1, b[0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 512; ++j) f[i][j] = b[j ^ a[i]] < a[i];
    for (int j = 0; j <= 512; ++j)
      if (f[i][j]) (b[j] = min(b[j], a[i])), ok[j] = 1;
  }
  int rp = 0;
  for (int i = 0; i <= 512; ++i)
    if (ok[i]) rp++;
  cout << rp << endl;
  for (int i = 0; i <= 512; ++i)
    if (ok[i]) printf("%d ", i);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
inline int read() {
  int x = 0, fh = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * fh;
}
inline void out(int *a, int l, int r) {
  for (int i = l; i <= r; ++i) printf("%d ", *(a + i));
  puts("");
}
const int N = 1e5 + 5, inf = 1e9;
int a[N], b[N], n, ok[N];
bool f[N][550];
signed main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 0; i <= 1024; ++i) b[i] = inf;
  ok[0] = 1, b[0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 512; ++j) f[i][j] = b[j ^ a[i]] < a[i];
    for (int j = 0; j <= 512; ++j)
      if (f[i][j]) (b[j] = min(b[j], a[i])), ok[j] = 1;
  }
  int rp = 0;
  for (int i = 0; i <= 512; ++i)
    if (ok[i]) rp++;
  cout << rp << endl;
  for (int i = 0; i <= 512; ++i)
    if (ok[i]) printf("%d ", i);
  return 0;
}

```