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 = 8192;
  const int maxn = 1e6 + 10;
  const int mod = 1e9 + 7;
  inline int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      x = (x << 1) + (x << 3) + c - '0';
      c = getchar();
    }
    return x * f;
  }
  vector<int> v[N];
  int n, m, a[maxn], b[N], st[N], rec[N], ans;
  int main() {
    n = read();
    ans = st[0] = 1;
    for (int i = 0; i < N; i++) v[i].push_back(0), rec[i] = N - 1;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      for (auto x : v[a[i]]) {
        int u = x ^ a[i];
        if (!st[u]) st[u] = 1, ++ans;
        if (a[i] <= rec[u]) {
          for (int j = a[i] + 1; j <= rec[u]; j++) v[j].push_back(u);
          rec[u] = a[i] - 1;
        }
      }
      v[a[i]].clear();
    }
    printf("%d\n", ans);
    for (int i = 0; i < N; i++)
      if (st[i]) printf("%d ", i);
    return 0;
  }
  