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>
  #pragma GCC target("avx2")
  #pragma GCC optimization("O3")
  #pragma GCC optimization("unroll-loops")
  using ll = long long;
  using ld = long double;
  using namespace std;
  const ll inf = int(2e9) + 1;
  int main() {
    ll n;
    cin >> n;
    vector<ll> s;
    map<int, vector<ll>> dd;
    for (int i = 0; i < n; i++) {
      int q;
      cin >> q;
      dd[q].push_back(i);
    }
    set<pair<int, int>> q;
    for (int i = 1; i <= 5000; i++) {
      for (int j = 0; j < 5; j++) {
        if (dd[i].size() == 0) continue;
        q.insert(make_pair(dd[i][(ll)round((dd[i].size() - 1) * j / 4)], i));
      }
    }
    for (auto el : q) s.push_back(el.second);
    ;
    n = s.size();
    vector<ll> used(10130, -inf);
    for (int i = n - 1; i >= 0; i--) {
      used[s[i]] = max(used[s[i]], s[i]);
      for (int j = 1; j <= 10120; j++) {
        if (used[j] > s[i]) {
          used[j ^ s[i]] = max(used[j ^ s[i]], s[i]);
        }
      }
    }
    ll res = 0;
    used[0] = 1;
    for (int i = 0; i <= 10120; i++)
      if (used[i] != -inf) res++;
    cout << res << endl;
    for (int i = 0; i <= 10120; i++)
      if (used[i] != -inf) cout << i << ' ';
    return 0;
  }
  