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 mx = 8192;
  int n, dp[10000], q[10000], p[10000], a[1000010], res;
  vector<int> v[5010];
  bool vis[10000];
  int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      v[a[i]].push_back(i);
    }
    dp[0] = 0;
    for (int i = 1; i <= mx; i++) dp[i] = n + 1, q[i] = i;
    for (int i = 1; i <= 5000; i++) {
      int tot = 0, now = -1;
      memset(vis, 0, sizeof(vis));
      for (int j = 0; j < v[i].size(); j++) {
        int x = v[i][j], lst = --tot;
        while (tot <= mx && dp[q[++tot]] < x)
          if (!vis[q[tot]]) p[++now] = q[tot];
        for (int j = lst; j < tot; j++) {
          if (dp[q[j] ^ i] > x) {
            p[++now] = q[j] ^ i;
            dp[p[now]] = x;
            vis[p[now]] = 1;
          }
        }
      }
      --tot;
      while (++tot <= mx)
        if (!vis[q[tot]]) p[++now] = q[tot];
      for (int i = 0; i <= mx; i++) q[i] = p[i];
    }
    for (int i = 0; i <= mx; i++)
      if (dp[i] <= n) res++;
    cout << res << endl;
    for (int i = 0; i <= mx; i++)
      if (dp[i] <= n) printf("%d ", i);
  }
  