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 = 1e6 + 11;
  int n, a[N], pos[N];
  vector<int> li[5011];
  int solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      li[a[i]].push_back(i);
    }
    for (int i = 0; i < 8192; i++) pos[i] = n + 1;
    pos[0] = 0;
    for (int x = 1; x <= 5000; x++) {
      if (li[x].size() == 0) continue;
      for (int y = 0; y < 8192; y++) {
        int z = x ^ y;
        if (pos[z] == n + 1 || pos[y] <= li[x][0]) {
          continue;
        }
        int id =
            lower_bound(li[x].begin(), li[x].end(), pos[z] + 1) - li[x].begin();
        if (id == li[x].size()) continue;
        pos[y] = min(pos[y], li[x][id]);
      }
    }
    int res = 0;
    for (int i = 0; i < 8192; i++)
      if (pos[i] <= n) res++;
    cout << res << '\n';
    for (int i = 0; i < 8192; i++)
      if (pos[i] <= n) cout << i << ' ';
    return 0;
  }
  int main() {
    int ntest = 1;
    while (ntest--) {
      solve();
    }
  }
  