#include <bits/stdc++.h>
using namespace std;
auto fio = (ios::sync_with_stdio(0), cin.tie(0), cout.tie(0));
int32_t main() {
  int n;
  cin >> n;
  vector<int> g(n);
  for (int& x : g) cin >> x;
  vector<vector<int>> idx(8192);
  for (int i = 0; i < n; i++) idx[g[i]].push_back(i);
  vector<int> mn(8192, n);
  mn[0] = -1;
  for (int i = 0; i <= 5000; i++) {
    if (idx[i].empty()) continue;
    for (int j = 0; j < 8192; j++) {
      if (mn[j] <= mn[i ^ j] || mn[i ^ j] >= idx[i].back()) continue;
      auto it = upper_bound(idx[i].begin(), idx[i].end(), mn[i ^ j]);
      if (it != idx[i].end() && *it < mn[j]) mn[j] = *it;
    }
  }
  cout << 8192 - count(mn.begin(), mn.end(), n) << endl;
  for (int i = 0; i < 8192; i++)
    if (mn[i] != n) cout << i << " ";
  cout << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
auto fio = (ios::sync_with_stdio(0), cin.tie(0), cout.tie(0));
int32_t main() {
  int n;
  cin >> n;
  vector<int> g(n);
  for (int& x : g) cin >> x;
  vector<vector<int>> idx(8192);
  for (int i = 0; i < n; i++) idx[g[i]].push_back(i);
  vector<int> mn(8192, n);
  mn[0] = -1;
  for (int i = 0; i <= 5000; i++) {
    if (idx[i].empty()) continue;
    for (int j = 0; j < 8192; j++) {
      if (mn[j] <= mn[i ^ j] || mn[i ^ j] >= idx[i].back()) continue;
      auto it = upper_bound(idx[i].begin(), idx[i].end(), mn[i ^ j]);
      if (it != idx[i].end() && *it < mn[j]) mn[j] = *it;
    }
  }
  cout << 8192 - count(mn.begin(), mn.end(), n) << endl;
  for (int i = 0; i < 8192; i++)
    if (mn[i] != n) cout << i << " ";
  cout << endl;
}

```