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;
  template <typename T>
  void chkMax(T &x, T y) {
    if (y > x) x = y;
  }
  template <typename T>
  void chkMin(T &x, T y) {
    if (y < x) x = y;
  }
  template <typename T>
  void inline read(T &x) {
    int f = 1;
    x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
  }
  const int N = 8200, INF = 0x3f3f3f3f;
  int n, a[N], g[N];
  vector<int> p[N];
  int main() {
    read(n);
    for (int i = 1, x; i <= n; i++) read(x), p[x].push_back(i);
    memset(g, 0x3f, sizeof g);
    g[0] = 0;
    for (int x = 1; x <= 5000; x++) {
      if (!p[x].size()) continue;
      for (int i = 0; i < N; i++) {
        if (g[i] == INF) continue;
        if (g[i] < p[x].back()) {
          int pt = upper_bound(p[x].begin(), p[x].end(), g[i]) - p[x].begin();
          int ps = p[x][pt];
          if (ps < g[i ^ x]) {
            g[i ^ x] = ps;
          }
        }
      }
    }
    int c = 0;
    for (int i = 0; i < N; i++)
      if (g[i] != INF) c++;
    printf("%d\n", c);
    for (int i = 0; i < N; i++)
      if (g[i] != INF) printf("%d ", i);
    return 0;
  }
  