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>
  int *ei[5000 + 1], eo[5000 + 1];
  void append(int a, int i) {
    int o = eo[a]++;
    if (o >= 2 && (o & o - 1) == 0)
      ei[a] = (int *)realloc(ei[a], o * 2 * sizeof *ei[a]);
    ei[a][o] = i;
  }
  int main() {
    static int ii[8192], bb[8192], bb_[8192];
    static char in[8192];
    int n, h, h_, i, a, b, o, cnt;
    scanf("%d", &n);
    for (a = 0; a <= 5000; a++) ei[a] = (int *)malloc(2 * sizeof *ei[a]);
    for (i = 0; i < n; i++) {
      scanf("%d", &a);
      append(a, i);
    }
    for (b = 0; b < 8192; b++) ii[b] = n, bb[b] = b;
    ii[0] = -1;
    for (a = 1; a <= 5000; a++) {
      memset(in, 0, 8192 * sizeof *in);
      cnt = 0;
      for (o = 0, h = 0; o <= eo[a]; o++, h = h_) {
        i = o == eo[a] ? n + 1 : ei[a][o];
        h_ = h;
        while (h_ < 8192 && ii[bb[h_]] < i) {
          if (!in[b = bb[h_]]) in[b] = 1, bb_[cnt++] = b;
          h_++;
        }
        h_ = h;
        while (h_ < 8192 && ii[bb[h_]] < i) {
          if (in[bb[h_]] == 1 && !in[b = bb[h_] ^ a])
            in[b] = 2, ii[b] = i, bb_[cnt++] = b;
          h_++;
        }
      }
      memcpy(bb, bb_, 8192 * sizeof *bb_);
    }
    cnt = 0;
    for (b = 0; b < 8192; b++)
      if (ii[b] != n) cnt++;
    printf("%d\n", cnt);
    for (b = 0; b < 8192; b++)
      if (ii[b] != n) printf("%d ", b);
    printf("\n");
    return 0;
  }
  