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 double PI = acos(-1.0);
  const long double EPS = 1e-7;
  const int MOD = 1e9 + 7;
  int POW(int b, int e) {
    int r = 1;
    while (e) {
      if (e % 2 == 1) {
        r *= b;
        r %= MOD;
      }
      e /= 2;
      b *= b;
      b %= MOD;
    }
    return r;
  }
  int gcd(int a, int b) {
    if (b > a) return gcd(b, a);
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  int INV(int a) { return POW(a, MOD - 2); }
  vector<vector<bool>> table(5001, vector<bool>(8192, false));
  vector<vector<int>> upd(5001);
  void setv(int a, int b) {
    if (!table[a][b]) {
      table[a][b] = true;
      if (a) {
        setv(a - 1, b);
      }
      upd[a].push_back(b);
    }
  }
  void roll(int a) {
    while (!upd[a].empty()) {
      setv(a - 1, upd[a].back() ^ a);
      upd[a].pop_back();
    }
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (int i = 0; i <= 5000; i++) {
      setv(i, 0);
    }
    for (int i = n - 1; i >= 0; i--) {
      if (a[i] > 0) roll(a[i]);
    }
    vector<int> out;
    for (int i = 0; i < 8192; i++) {
      if (table[0][i]) {
        out.push_back(i);
      }
    }
    cout << (int)out.size() << '\n';
    for (auto v : out) {
      cout << v << ' ';
    }
  }
  