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>
  using MinHeap = priority_queue<T, vector<T>, greater<T>>;
  template <typename T>
  using MaxHeap = priority_queue<T>;
  const long double PI = 3.14159265358979323846;
  const long double E = 2.718281828459;
  const long double EPS = 1e-8;
  long long MOD = 1e9 + 7;
  long long M(long long n) { return (n % MOD + MOD) % MOD; }
  long long pow2(long long n) { return 1ll << n; }
  mt19937 rng(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  const long long MX = 5e3, MASK = 8192;
  set<long long> se[MX + 2];
  int main(int argn, char** argv) {
    ios::sync_with_stdio(false);
    cout.tie(nullptr);
    cin.tie(nullptr);
    int T = 1;
    for (long long(I) = (0); (I) <= ((T)-1); (I)++) {
      long long n;
      cin >> n;
      vector<long long> a(n);
      for (long long(i) = (0); (i) <= ((n)-1); (i)++) cin >> a[i];
      for (long long(i) = (0); (i) <= ((n)-1); (i)++) se[a[i]].emplace(i);
      long long dp[MASK];
      memset(dp, 0x7f, sizeof(dp));
      dp[0] = 0;
      for (long long(i) = (1); (i) <= (MX); (i)++) {
        set<long long>& s = se[i];
        for (long long(mask) = (0); (mask) <= ((MASK)-1); (mask)++) {
          auto iter = s.lower_bound(dp[mask]);
          if (iter != s.end()) dp[mask ^ i] = min(dp[mask ^ i], *iter);
        }
      }
      vector<long long> ans;
      for (long long(x) = (0); (x) <= ((MASK)-1); (x)++)
        if (dp[x] < 1e6) ans.emplace_back(x);
      long long sz = ans.size();
      cout << sz << endl;
      for (auto x : ans) cout << x << " ";
      cout << endl;
    }
    return 0;
  }
  