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 = (int)1e6 + 50, M = 8200;
  int n, a[N];
  vector<int> xs[M];
  vector<pair<int, int> > ear;
  bool vis[M];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n;
    for (int i = 0; i < (n); ++i) cin >> a[i], xs[a[i]].push_back(i);
    ear.push_back({0, -1});
    for (int v = 0; v < (M); ++v) {
      fill(vis, vis + M, 0);
      vector<pair<int, int> > gear;
      int c = 0;
      for (int i : xs[v]) {
        int r = c;
        while (r < (int)(ear).size() && ear[r].second < i) {
          if (!vis[ear[r].first]) {
            gear.push_back({ear[r]});
            vis[ear[r].first] = true;
          }
          r++;
        }
        for (int j = c; j < r; j++) {
          int nxt = ear[j].first ^ v;
          if (!vis[nxt]) {
            gear.push_back({nxt, i});
            vis[nxt] = true;
          }
        }
        c = r;
      }
      while (c < (int)(ear).size()) {
        if (!vis[ear[c].first]) {
          gear.push_back({ear[c]});
          vis[ear[c].first] = true;
        }
        c++;
      }
      ear = gear;
    }
    sort(begin(ear), end(ear));
    cout << (int)(ear).size() << '\n';
    for (int i = 0; i < ((int)(ear).size()); ++i) cout << ear[i].first << " ";
    cout << endl;
  }
  