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;
  const int A = 5e3 + 5, K = 1 << 13;
  int n, dp[A][K];
  vector<int> pos[A];
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    cin >> n;
    for (int i = 0; i < (A); ++i) {
      for (int j = 0; j < (K); ++j) {
        dp[i][j] = n;
      }
    }
    dp[0][0] = -1;
    for (int i = 0; i < (n); ++i) {
      int a;
      cin >> a;
      pos[a].push_back(i);
    }
    for (int i = 1; i < (A); ++i) {
      for (int j = 0; j < (K); ++j) {
        dp[i][j] = dp[i - 1][j];
        auto it = lower_bound(begin(pos[i]), end(pos[i]), dp[i - 1][j ^ i]);
        if (it != end(pos[i])) {
          dp[i][j] = min(dp[i][j], *it);
        }
      }
    }
    vector<int> ans;
    for (int j = 0; j < (K); ++j) {
      if (dp[A - 1][j] < n) {
        ans.push_back(j);
      }
    }
    cout << (int)(ans).size() << '\n';
    for (const int x : ans) {
      cout << x << ' ';
    }
    cout << '\n';
  }
  