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 N = 1000000 + 9;
  int a[N], nxt[N], pos[N], vis[N], minv[N];
  vector<int> vc[N];
  void solve() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", a + i);
    }
    for (int i = n; i >= 1; --i) {
      nxt[i] = pos[a[i]];
      pos[a[i]] = i;
    }
    minv[0] = 0;
    for (int i = 1; i < (1 << 13); ++i) {
      minv[i] = 5000;
    }
    vis[0] = true;
    for (int i = 1; i <= 5000; ++i) {
      if (pos[i]) vc[pos[i]].push_back(0);
    }
    for (int i = 1; i <= n; ++i) {
      for (int j : vc[i]) {
        int k = j ^ a[i];
        vis[k] = true;
        if (a[i] < minv[k]) {
          for (int x = a[i] + 1; x <= minv[k]; ++x) {
            if (pos[x]) vc[pos[x]].push_back(k);
          }
          minv[k] = a[i];
        }
      }
      pos[a[i]] = nxt[i];
    }
    vector<int> ans;
    for (int i = 0; i < (1 << 13); ++i) {
      if (vis[i]) ans.push_back(i);
    }
    printf("%d\n", (int)ans.size());
    for (int i : ans) printf("%d ", i);
    printf("\n");
  }
  int main() {
    int T = 1;
    while (T--) solve();
    return 0;
  }
  