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 long long inf = 1e18;
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie();
    cout.tie();
    long long n;
    cin >> n;
    vector<long long> a(n);
    map<long long, long long> mp;
    vector<long long> cnt(n);
    for (long long i = 0; i < n; ++i) {
      long long o;
      cin >> o;
      a[i] = o;
      mp[o]++;
      cnt[i] = mp[o];
    }
    vector<long long> n_a;
    for (long long i1 = 0; i1 < n; ++i1) {
      if (cnt[i1] == 1 || cnt[i1] == mp[a[i1]] || cnt[i1] == 2 ||
          cnt[i1] == mp[a[i1]] - 1) {
        n_a.push_back(a[i1]);
      }
    }
    a = n_a;
    n = a.size();
    long long mx = 8191;
    bool dp[n + 1][mx + 1];
    for (long long j = 0; j < n + 1; ++j) {
      for (long long i = 0; i < mx + 1; ++i) {
        dp[j][i] = false;
      }
    }
    dp[0][0] = true;
    long long last[mx + 1];
    for (long long m = 0; m < mx + 1; ++m) {
      last[m] = inf;
    }
    last[0] = -inf;
    for (long long k = 1; k < n + 1; ++k) {
      for (long long i = 0; i < mx + 1; ++i) {
        long long need = i ^ a[k - 1];
        if (last[need] < a[k - 1]) {
          dp[k][i] = true;
        }
      }
      for (long long j = 0; j < mx + 1; ++j) {
        if (dp[k][j]) last[j] = min(last[j], a[k - 1]);
      }
    }
    vector<bool> used(mx + 1);
    for (long long k1 = 0; k1 <= n; ++k1) {
      for (long long l = 0; l < mx + 1; ++l) {
        if (dp[k1][l]) used[l] = true;
      }
    }
    long long cc = 0;
    for (long long l1 = 0; l1 < mx + 1; ++l1) {
      cc += used[l1];
    }
    cout << cc << "\n";
    for (long long l1 = 0; l1 < mx + 1; ++l1) {
      if (used[l1]) {
        cout << l1 << " ";
      }
    }
    cout << "\n";
  }
  