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;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  void _do(T x) {
    cerr << x << "\n";
  }
  template <typename T, typename... U>
  void _do(T x, U... y) {
    cerr << x << ", ";
    _do(y...);
  }
  const int MOD1 = 1e9 + 7;
  const int MOD2 = 998244353;
  const long long INF = 3e18;
  const long double PI = 3.14159265358979323846;
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long fpow(long long a, long long b, long long m) {
    if (!b) return 1;
    long long ans = fpow(a * a % m, b / 2, m);
    return (b % 2 ? ans * a % m : ans);
  }
  long long inv(long long a, long long m) { return fpow(a, m - 2, m); }
  const long long K = 8192;
  long long a[1000005], cur[K];
  vector<long long> ed[5005];
  bool can[K];
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long n;
    cin >> n;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    for (long long i = 0; i < 5001; i++) ed[i].push_back(0);
    for (long long i = 0; i < K; i++) cur[i] = 5001;
    can[0] = 1;
    for (long long i = 1; i <= n; i++) {
      for (long long x : ed[a[i]]) {
        long long y = x ^ a[i];
        can[y] = 1;
        while (cur[y] > a[i]) {
          cur[y]--;
          if (cur[y] != a[i]) ed[cur[y]].push_back(y);
        }
      }
      ed[a[i]].clear();
    }
    vector<long long> ans;
    for (long long i = 0; i < K; i++)
      if (can[i]) ans.push_back(i);
    cout << ans.size() << "\n";
    for (long long u : ans) cout << u << " ";
  }
  