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;
  void debug_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << H;
    debug_out(T...);
  }
  constexpr int MAXN = 1e6 + 5;
  constexpr int MOD = 1e9 + 7;
  constexpr int INF = 1e9 + 1;
  constexpr int LOG = 3e1 + 1;
  constexpr int SQ = 500;
  constexpr int ALP = 5e3 + 3;
  int n;
  int a[MAXN];
  bool mark[ALP][ALP * 2], has[ALP * 2];
  vector<int> G[ALP], ans;
  void add(int idx, int x) {
    for (int i = idx + 1; i <= 5000; i++) {
      if (!mark[i][x]) {
        G[i].push_back(x);
        mark[i][x] = true;
      } else {
        break;
      }
    }
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (int i = 1; i <= 5000; i++) {
      G[i].push_back(0);
      mark[i][0] = true;
    }
    ans.push_back(0);
    has[0] = true;
    for (int i = 1; i <= n; i++) {
      int ptr = 0;
      while (ptr < G[a[i]].size()) {
        int x = G[a[i]][ptr++] ^ a[i];
        if (!has[x]) {
          ans.push_back(x);
          has[x] = true;
        }
        add(a[i], x);
      }
      G[a[i]].clear();
    }
    sort(ans.begin(), ans.end());
    cout << ans.size() << "\n";
    for (auto x : ans) {
      cout << x << " ";
    }
    return 0;
  }
  