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>
  const int maxn = 1e6 + 10;
  const int maxm = (1 << 13);
  std::vector<int> g[maxm];
  int r[maxm];
  int n;
  int ans[maxm];
  int main() {
    scanf("%d", &n);
    ans[0] = 1;
    for (int i = 1; i < maxm; i++) {
      g[i].emplace_back(0);
    }
    for (int i = 1; i < maxm; i++) {
      r[i] = maxm;
    }
    for (int i = 1; i <= n; i++) {
      int val;
      scanf("%d", &val);
      for (const auto &pre : g[val]) {
        int nk = (pre ^ val);
        ans[nk] = 1;
        while (r[nk] > val) {
          --r[nk];
          if (r[nk] != val) g[r[nk]].emplace_back(nk);
        }
      }
      g[val].clear();
    }
    std::vector<int> vec;
    for (int i = 0; i < maxm; i++) {
      if (ans[i]) vec.emplace_back(i);
    }
    int lens = vec.size();
    printf("%d\n", lens);
    for (int i = 0; i < lens; i++) {
      printf("%d%c", vec[i], " \n"[i == lens - 1]);
    }
    return 0;
  }
  