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 int oo = 1e9;
  const int N = 5001;
  const int maxxor = (1 << 13);
  int f[N][maxxor];
  vector<int> a[N];
  int cnp(int l, int r, int pos, int id) {
    if (l > r) return oo;
    int mid = (l + r) / 2;
    if (a[id][mid] > pos) return min(a[id][mid], cnp(l, mid - 1, pos, id));
    return cnp(mid + 1, r, pos, id);
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      a[x].push_back(i);
    }
    for (int i = 0; i < N; i++)
      for (int j = 0; j < maxxor; j++) f[i][j] = oo;
    f[0][0] = 0;
    int id;
    vector<int> res;
    for (int i = 1; i < N; i++)
      for (int j = 0; j < maxxor; j++) {
        f[i][j] = f[i - 1][j];
        if (a[i].size() == 0) {
          if (i == N - 1 && f[i][j] != oo) res.push_back(j);
          continue;
        }
        int pos = f[i - 1][j ^ i];
        id = cnp(0, a[i].size() - 1, pos, i);
        if (id != oo) f[i][j] = min(f[i][j], id);
        if (i == N - 1 && f[i][j] != oo) res.push_back(j);
      }
    cout << res.size() << endl;
    for (int xnew : res) cout << xnew << " ";
    return 0;
  }
  