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("O3")
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  long long arr[1010100];
  bool cmp(pair<long long, long long> p1, pair<long long, long long> p2) {
    if (p1.first != p2.first)
      return p1.first < p2.first;
    else
      return p1.second > p2.second;
  }
  void solve() {
    long long N;
    cin >> N;
    long long i;
    vector<vector<long long>> v(8192);
    vector<long long> chk(8192, 1000000000000);
    vector<long long> ans(8192);
    for (i = 1; i <= N; i++) cin >> arr[i], v[arr[i]].push_back(i);
    chk[0] = 0;
    v[0].push_back(0);
    for (i = 0; i <= 5000; i++) {
      long long j;
      if (v[i].empty()) continue;
      vector<long long> nw(8192, 1000000000000);
      nw[0] = 0;
      for (j = 0; j < 8192; j++) {
        long long x = i ^ j;
        long long ind =
            upper_bound(v[i].begin(), v[i].end(), chk[x]) - v[i].begin();
        if (ind == v[i].size()) continue;
        ans[j] = 1;
        nw[j] = min(nw[j], v[i][ind]);
      }
      for (j = 0; j < 8192; j++) chk[j] = min(chk[j], nw[j]);
    }
    long long cnt = 0;
    ans[0] = 1;
    for (i = 0; i < 8192; i++) cnt += ans[i];
    cout << cnt << '\n';
    for (i = 0; i < 8192; i++)
      if (ans[i]) cout << i << ' ';
    cout << '\n';
  }
  void init() {}
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0);
    long long T;
    T = 1;
    while (T--) {
      init();
      solve();
    }
  }
  