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>
  const int N = 1000005;
  const int M = 10000;
  int n, t;
  inline void read(int &x) {
    register char ch = 0;
    register bool w = 0;
    for (x = 0; !std::isdigit(ch); w |= ch == '-', ch = getchar())
      ;
    for (; std::isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
      ;
    (w) && (x = -x);
    return;
  }
  std::vector<int> reach[M];
  int min0[M], vis[M], res[M];
  void solve() {
    read(n);
    static int ai[N];
    for (int i = 1; i <= n; ++i) read(ai[i]);
    for (int i = 0; i < 8192; ++i) reach[i].push_back(0), min0[i] = 8192;
    for (int i = 1; i <= n; ++i) {
      if (vis[ai[i]]) continue;
      vis[ai[i]] = 1;
      for (auto v : reach[ai[i]]) {
        int xr = v ^ ai[i];
        for (int j = ai[i] + 1; j < min0[xr]; ++j) {
          if (vis[j]) std::vector<int>().swap(reach[j]), vis[j] = 0;
          reach[j].push_back(xr);
        }
        res[xr] = 1;
        min0[xr] = std::min(min0[xr], ai[i] + 1);
      }
    }
    res[0] = 1;
    std::vector<int> vec;
    for (int i = 0; i < 8192; ++i)
      if (res[i]) vec.push_back(i);
    printf("%lu\n", vec.size());
    for (auto v : vec) printf("%d ", v);
    printf("\n");
  }
  int main() {
    t = 1;
    while (t--) solve();
    return 0;
  }
  