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 solve();
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    solve();
    return 0;
  }
  const int N = 1e6 + 100, K = 10001, INF = 2e18, MOD = 1e9 + 7;
  int n, a[N];
  vector<bool> ans(K);
  vector<vector<bool>> ok(K, vector<bool>(K));
  vector<int> v[K];
  void solve() {
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (int i = 0; i < K; i++) {
      ok[i][0] = true;
      v[i].push_back(0);
    }
    for (int i = 0; i < n; i++) {
      for (int x : v[a[i]]) {
        int e = (x ^ a[i]);
        for (int h = a[i] + 1; h < K; h++) {
          if (ok[h][e]) break;
          if (!ok[h][e]) {
            ok[h][e] = true;
            v[h].push_back(e);
          }
        }
      }
      v[a[i]].clear();
    }
    for (int i = 0; i < K; i++) {
      for (int j = 0; j < K; j++) {
        if (ok[i][j]) {
          ans[j] = true;
        }
      }
    }
    int cnt = 0;
    for (int i = 0; i < K; i++) {
      cnt += ans[i];
    }
    cout << cnt << '\n';
    for (int i = 0; i < K; i++) {
      if (ans[i]) {
        cout << i << " ";
      }
    }
  }
  