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;
  inline long long md(long long m, long long v) { return ((v % m) + m) % m; }
  const int MAX = 8200;
  const int NOT = 1e9;
  void solve() {
    int n;
    cin >> n;
    vector<int> arr(n);
    for (int i = 0; i < n; i++) cin >> arr[i];
    vector<vector<int> > idx(MAX + 1);
    for (int i = 0; i < n; i++) {
      idx[arr[i]].push_back(i);
    }
    vector<int> dp(MAX + 1, NOT);
    dp[0] = -1;
    for (int a = 1; a <= MAX; a++) {
      int I = a % 2, J = 1 - I;
      if ((int)(idx[a]).size() == 0) continue;
      int upper = idx[a].back();
      dp[a] = min(dp[a], idx[a].front());
      for (int b = 1; b <= MAX; b++) {
        int target = a ^ b;
        if (target > MAX) continue;
        if (dp[target] < upper) {
          int where =
              int(lower_bound((idx[a]).begin(), (idx[a]).end(), dp[target]) -
                  begin(idx[a]));
          dp[b] = min(dp[b], idx[a][where]);
        }
      }
    }
    vector<int> ans;
    for (int i = 0; i <= MAX; i++) {
      if (dp[i] != NOT) ans.push_back(i);
    }
    cout << (int)(ans).size() << "\n";
    for (int i : ans) cout << i << ' ';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    solve();
    return 0;
  }
  