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 read(T &x) {
    x = 0;
    char c = getchar();
    T sig = 1;
    for (; !isdigit(c); c = getchar())
      if (c == '-') sig = -1;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= sig;
  }
  int dp[5001][8192]{};
  class Solution {
   public:
    void solve() {
      int n;
      read(n);
      vector<int> a(n);
      vector<vector<int>> nums(5001);
      for (int i = 0; i < n; ++i) {
        read(a[i]);
        nums[a[i]].push_back(i + 1);
      }
      memset(dp, 0x3f, sizeof(dp));
      dp[0][0] = 0;
      vector<bool> can(8192);
      can[0] = true;
      for (int i = 1; i <= 5000; ++i) {
        for (int j = 1; j < 8192; ++j) {
          dp[i][j] = dp[i - 1][j];
          if (i == j && !nums[i].empty()) dp[i][j] = min(dp[i][j], nums[i][0]);
          auto it = upper_bound(nums[i].begin(), nums[i].end(), dp[i - 1][i ^ j]);
          if (it != nums[i].end()) dp[i][j] = min(dp[i][j], *it);
          if (dp[i][j] < 0x3f3f3f3f) can[j] = true;
        }
      }
      int ans = 0;
      for (int i = 0; i < 8192; ++i)
        if (can[i]) ans++;
      printf("%d\n", ans);
      for (int i = 0; i < 8192; ++i)
        if (can[i]) printf("%d ", i);
      printf("\n");
    }
  };
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    Solution solution = Solution();
    solution.solve();
  }
  