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;
  std::mt19937 rnd(
      (int)std::chrono::steady_clock::now().time_since_epoch().count());
  long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
  const int MAXN = 1000000;
  const int NBIT = 13;
  const int MAXA = 5000;
  int n;
  int a[MAXN];
  vector<int> ans;
  int best[1 << NBIT];
  vector<int> opt[MAXA + 1];
  void solve() {
    for (int i = (0); i < (1 << NBIT); ++i) best[i] = INT_MAX;
    auto relax = [&](int at, int mx) {
      if (mx >= best[at]) return;
      for (int i = (mx + 1); i <= (min(best[at], MAXA)); ++i)
        opt[i].push_back(at);
      best[at] = mx;
    };
    relax(0, -1);
    for (int i = (0); i < (n); ++i) {
      int cur = a[i];
      while (((int)(opt[cur]).size()) > 0) {
        int at = opt[cur].back();
        opt[cur].pop_back();
        assert(best[at] < cur);
        relax(at ^ cur, cur);
      }
    }
    ans.clear();
    for (int i = (0); i < (1 << NBIT); ++i)
      if (best[i] != INT_MAX) ans.push_back(i);
  }
  void run() {
    scanf("%d", &n);
    for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
    solve();
    printf("%d\n", ((int)(ans).size()));
    for (int i = (0); i < (((int)(ans).size())); ++i) {
      if (i != 0) printf(" ");
      printf("%d", ans[i]);
    }
    puts("");
  }
  int main() {
    run();
    return 0;
  }
  