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;
  int n, x, g, w, num, now, p[1000100], s[1000100], b[1000100];
  pair<int, int> a[1000100];
  int main() {
    for (int i = 1; i <= 8191; ++i) p[i] = 10000000;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i].first);
      a[i].second = i;
    }
    sort(a + 1, a + 1 + n);
    now = 1;
    while (now <= n) {
      w = 1;
      g = a[now].first;
      b[w] = a[now].second;
      now++;
      while (now <= n && a[now].first == g) b[++w] = a[now].second, now++;
      p[g] = min(p[g], b[1]);
      for (int i = 1; i <= 8191; ++i)
        if (p[i] < b[w])
          p[i ^ g] = min(p[i ^ g], b[lower_bound(b + 1, b + 1 + w, p[i]) - b]);
    }
    for (int i = 1; i <= 8191; ++i)
      if (p[i] <= 1000000) num++;
    printf("%d\n0", num + 1);
    for (int i = 1; i <= 8191; ++i)
      if (p[i] <= 1000000) printf(" %d", i);
    return 0;
  }
  