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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  char Buf[1 << 23], *P1 = Buf, *P2 = Buf;
  template <typename Tp = int>
  inline Tp read() {
    Tp x = 0;
    bool f = false;
    char ch =
        (P1 == P2 && (P2 = (P1 = Buf) + fread(Buf, 1, 1 << 23, stdin), P1 == P2)
             ? EOF
             : *P1++);
    while (ch < '0' || ch > '9')
      f |= (ch == '-'),
          ch = (P1 == P2 && (P2 = (P1 = Buf) + fread(Buf, 1, 1 << 23, stdin),
                             P1 == P2)
                    ? EOF
                    : *P1++);
    while (ch >= '0' && ch <= '9')
      x = x * 10 + (ch ^ 48),
      ch =
          (P1 == P2 && (P2 = (P1 = Buf) + fread(Buf, 1, 1 << 23, stdin), P1 == P2)
               ? EOF
               : *P1++);
    return f ? -x : x;
  }
  template <typename Tp>
  inline void read(Tp &x) {
    x = read<Tp>();
  }
  template <typename Tp1, typename... Tp2>
  inline void read(Tp1 &x, Tp2 &...y) {
    read(x), read(y...);
  }
  constexpr int Max = 1e5 + 100;
  constexpr int max_num = 8192;
  int n, x;
  bool can[max_num];
  vector<int> f[max_num];
  int pre[max_num];
  signed main() {
    read(n);
    for (int i = 0; i < max_num; ++i) f[i].emplace_back(0), pre[i] = max_num;
    can[0] = 1;
    while (n--) {
      int i;
      read(i);
      for (int x : f[i]) {
        const int y = x ^ i;
        can[y] = true;
        while (pre[y] > i) {
          --pre[y];
          if (pre[y] > i) f[pre[y]].emplace_back(y);
        }
      }
      f[i].clear();
    }
    int ans = 0;
    for (int i = 0; i < max_num; ++i) ans += can[i];
    printf("%d\n", ans);
    for (int i = 0; i < max_num; ++i)
      if (can[i]) printf("%d ", i);
    putchar('\n');
    return 0;
  }
  