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>
  const int N = 1000100;
  int n, k, a, dp[2][N], ans[N], cur;
  std::vector<int> v[N];
  int main() {
    scanf("%d", &n);
    for (int i = 0; i < 8192; ++i) dp[cur][i] = dp[cur ^ 1][i] = n + 1;
    dp[cur][0] = 0, k = 0;
    for (int i = 1; i <= n; ++i) scanf("%d", &a), v[a].push_back(i);
    for (int i = 0; i < 5000; ++i, cur ^= 1) {
      for (int j = 0; j < 8192; ++j) dp[cur ^ 1][j] = dp[cur][j];
      if (!v[i + 1].size()) continue;
      for (int j = 0; j < 8192; ++j) {
        if (dp[cur][j] == n + 1) continue;
        int id = std::lower_bound(v[i + 1].begin(), v[i + 1].end(), dp[cur][j]) -
                 v[i + 1].begin();
        if (id == v[i + 1].size()) continue;
        dp[cur ^ 1][j ^ (i + 1)] =
            std::min(dp[cur ^ 1][j ^ (i + 1)], v[i + 1][id]);
      }
    }
    for (int i = 0; i < 8192; ++i)
      if (dp[cur][i] != n + 1) ans[++k] = i;
    printf("%d\n", k);
    for (int i = 1; i <= k; ++i) printf("%d ", ans[i]);
    printf("\n");
    return 0;
  }
  