description: 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] 
solution: #include <bits/stdc++.h>
  using namespace std;
  const int N = 5000 + 10;
  bool vis[N][1 << 13];
  int t[1000 * 1000 + 10];
  vector<int> l[N];
  bool res[1 << 13];
  void solve() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> t[i];
    for (int i = 1; i <= 5000; i++) l[i].push_back(0), vis[i][0] = 1;
    res[0] = 1;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < l[t[i]].size(); j++) {
        for (int k = t[i] + 1; k <= 5000; k++) {
          if (vis[k][l[t[i]][j] ^ t[i]]) break;
          vis[k][l[t[i]][j] ^ t[i]] = 1;
          l[k].push_back(l[t[i]][j] ^ t[i]);
        }
        res[l[t[i]][j] ^ t[i]] = 1;
      }
      l[t[i]].clear();
    }
    vector<int> ans;
    for (int i = 0; i < (1 << 13); i++)
      if (res[i]) ans.push_back(i);
    cout << ans.size() << "\n";
    for (auto i : ans) cout << i << " ";
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    solve();
  }
  