#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tt = 1;
  while (tt--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    int max_val = (1 << 13);
    int od[max_val];
    vector<int> vals[max_val];
    for (int i = 0; i < max_val; i++) od[i] = max_val;
    for (int i = 0; i < max_val; i++) vals[i].push_back(0);
    vector<int> pos(max_val, 0);
    pos[0] = 1;
    for (int i = 0; i < n; i++) {
      int cur = a[i];
      for (int x : vals[cur]) {
        int res = x ^ cur;
        pos[res] = 1;
        while (od[res] > cur) {
          od[res]--;
          if (od[res] != cur) vals[od[res]].push_back(res);
        }
      }
      vals[cur].clear();
    }
    int cnt = 0;
    for (int i = 0; i < max_val; i++) cnt += pos[i];
    cout << cnt << "\n";
    for (int i = 0; i < max_val; i++) {
      if (pos[i]) {
        cout << i << " ";
      }
    }
  }
}

### Prompt

Please formulate a cpp solution to the following problem:

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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tt = 1;
  while (tt--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    int max_val = (1 << 13);
    int od[max_val];
    vector<int> vals[max_val];
    for (int i = 0; i < max_val; i++) od[i] = max_val;
    for (int i = 0; i < max_val; i++) vals[i].push_back(0);
    vector<int> pos(max_val, 0);
    pos[0] = 1;
    for (int i = 0; i < n; i++) {
      int cur = a[i];
      for (int x : vals[cur]) {
        int res = x ^ cur;
        pos[res] = 1;
        while (od[res] > cur) {
          od[res]--;
          if (od[res] != cur) vals[od[res]].push_back(res);
        }
      }
      vals[cur].clear();
    }
    int cnt = 0;
    for (int i = 0; i < max_val; i++) cnt += pos[i];
    cout << cnt << "\n";
    for (int i = 0; i < max_val; i++) {
      if (pos[i]) {
        cout << i << " ";
      }
    }
  }
}

```