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;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long long INF = 1e18 + 123;
  const long double EPS = 1e-9;
  const int inf = 1e9 + 123;
  const int MOD = 1e9 + 7;
  const int N = 1e5 + 123;
  const int M = 1e6 + 1;
  const int K = 8192;
  const int Q = 5001;
  const double pi = 3.14159265359;
  const int dx[] = {0, 0, 1, -1};
  const int dy[] = {1, -1, 0, 0};
  bool was[K][Q], ok[K];
  vector<int> v[N];
  void solve() {
    int n;
    cin >> n;
    ok[0] = 1;
    for (int second = 1; second < Q; second++) {
      v[second].push_back(0);
      was[0][second] = 1;
    }
    for (int i = 1, a; i <= n; i++) {
      cin >> a;
      for (int first : v[a]) {
        ok[first ^ a] = 1;
        for (int second = a + 1; second < Q && !was[first ^ a][second];
             second++) {
          was[first ^ a][second] = 1;
          v[second].push_back(first ^ a);
        }
      }
      v[a].clear();
    }
    vector<int> v;
    for (int first = 0; first < K; first++) {
      if (ok[first]) {
        v.push_back(first);
      }
    }
    cout << (int)v.size() << '\n';
    for (int first : v) cout << first << " ";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    for (int i = 1; i <= t; i++) {
      solve();
    }
    return 0;
  }
  