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 optimize("Ofast")
  using namespace std;
  const int N = (2 << 13);
  int can[N];
  vector<int> q[N];
  int mn[N];
  void Set(int p, int j) {
    if (p >= mn[j]) return;
    for (int tp = p; tp < mn[j]; tp++) {
      if (tp + 1 < N && mn[j ^ (tp + 1)] > tp + 1) {
        q[tp + 1].push_back(j ^ (tp + 1));
      }
    }
    mn[j] = p;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    fill(mn, mn + N, N);
    int n;
    cin >> n;
    Set(0, 0);
    while (n--) {
      int p;
      cin >> p;
      for (auto j : q[p]) {
        Set(p, j);
      }
      q[p].clear();
    }
    int cnt = 0;
    for (int j = 0; j < N; j++) {
      can[j] = (mn[j] != N);
      cnt += can[j];
    }
    cout << cnt << '\n';
    for (int j = 0; j < N; j++) {
      if (can[j]) {
        cout << j << ' ';
      }
    }
    cout << '\n';
  }
  