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 = 5e3 + 10, mod = 998244353;
  vector<int> v[N];
  int dp[N][8192];
  int main() {
    int n;
    scanf("%d", &n);
    memset(dp, 0x3f, sizeof dp);
    dp[0][0] = 0;
    for (int i = 1, x; i <= n; i++) scanf("%d", &x), v[x].push_back(i);
    for (int i = 1; i <= 5e3; i++)
      for (int j = 0; j <= 8191; j++) {
        dp[i][j] = dp[i - 1][j];
        if ((j ^ i) > 8191 || !v[i].size() || v[i].back() < dp[i - 1][j ^ i])
          continue;
        dp[i][j] = min(dp[i][j],
                       *lower_bound(v[i].begin(), v[i].end(), dp[i - 1][j ^ i]));
      }
    for (int i = 0; i <= 8191; i++)
      if (dp[5000][i] != 0x3f3f3f3f) v[5001].push_back(i);
    cout << v[5001].size() << endl;
    for (auto x : v[5001]) cout << x << ' ';
    return 0;
  }
  