#include <bits/stdc++.h>
using namespace std;
const int maxx = (1 << 13);
int main() {
  ios::sync_with_stdio(false);
  vector<int> ans(maxx, 0);
  vector<vector<int>> g(maxx, vector<int>());
  vector<vector<int>> used(maxx, vector<int>(maxx, 0));
  ans[0] = 1;
  for (int i = 0; i < maxx; i++) {
    used[0][i] = 1;
    g[i].push_back(0);
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    for (int to : g[x]) {
      int val = x ^ to;
      ans[val] = 1;
      for (int j = x + 1; j < maxx && !used[val][j]; j++) {
        used[val][j] = 1;
        g[j].push_back(val);
      }
    }
    g[x].clear();
  }
  cout << accumulate(begin(ans), end(ans), 0) << endl;
  for (int i = 0; i < maxx; i++) {
    if (ans[i]) {
      cout << i << " ";
    }
  }
  cout << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

This is a harder version of the problem with bigger 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 ≤ 10^6).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 5000) — 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;
const int maxx = (1 << 13);
int main() {
  ios::sync_with_stdio(false);
  vector<int> ans(maxx, 0);
  vector<vector<int>> g(maxx, vector<int>());
  vector<vector<int>> used(maxx, vector<int>(maxx, 0));
  ans[0] = 1;
  for (int i = 0; i < maxx; i++) {
    used[0][i] = 1;
    g[i].push_back(0);
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    for (int to : g[x]) {
      int val = x ^ to;
      ans[val] = 1;
      for (int j = x + 1; j < maxx && !used[val][j]; j++) {
        used[val][j] = 1;
        g[j].push_back(val);
      }
    }
    g[x].clear();
  }
  cout << accumulate(begin(ans), end(ans), 0) << endl;
  for (int i = 0; i < maxx; i++) {
    if (ans[i]) {
      cout << i << " ";
    }
  }
  cout << endl;
  return 0;
}

```