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;
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie();
    cout.tie();
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
      int o;
      cin >> o;
      a[i] = o;
    }
    int mx = 8191;
    vector<bool> can[mx + 1];
    vector<bool> was[mx + 1];
    for (int j = 0; j < mx + 1; ++j) {
      can[j].resize(mx + 1);
      was[j].resize(mx + 1);
    }
    vector<int> cc[mx + 1];
    cc[0].push_back(0);
    for (int m = 0; m < n; ++m) {
      cc[a[m]].push_back(0);
      can[a[m]][0] = true;
    }
    for (int k = 0; k < n; ++k) {
      for (int i = 0; i < cc[a[k]].size(); ++i) {
        int f = cc[a[k]][i] ^ a[k];
        for (int j = a[k] + 1; j < mx + 1; ++j) {
          if (can[j][f] || was[j][f]) break;
          cc[j].push_back(f);
          can[j][f] = true;
        }
        was[a[k]][cc[a[k]][i]] = true;
      }
      cc[a[k]].clear();
    }
    vector<bool> used(mx + 1);
    for (int k1 = 0; k1 < mx + 1; ++k1) {
      for (int l = 0; l < mx + 1; ++l) {
        if (can[k1][l]) used[l] = true;
      }
    }
    int ccc = 0;
    for (int l1 = 0; l1 < mx + 1; ++l1) {
      ccc += used[l1];
    }
    cout << ccc << "\n";
    for (int l1 = 0; l1 < mx + 1; ++l1) {
      if (used[l1]) {
        cout << l1 << " ";
      }
    }
    cout << "\n";
  }
  