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 kMax = 1 << 13;
  int dp[kMax];
  vector<int> pos[kMax];
  void min_self(int &x, const int &y) {
    if (y < x) {
      x = y;
    }
  }
  void TestCase() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
      int x;
      cin >> x;
      pos[x].emplace_back(i);
    }
    memset(dp, 0x3f, sizeof(dp));
    dp[0] = -1;
    for (int x = 0; x < kMax; ++x) {
      if (!pos[x].empty()) {
        for (int sxor = 0; sxor < kMax; ++sxor) {
          if (dp[sxor] != 0x3f3f3f3f) {
            auto it = upper_bound(pos[x].begin(), pos[x].end(), dp[sxor]);
            if (it != pos[x].end()) {
              min_self(dp[sxor ^ x], *it);
            }
          }
        }
      }
    }
    int ans = 0;
    for (int x = 0; x < (1 << 13); ++x) {
      ans += (dp[x] != 0x3f3f3f3f);
    }
    cout << ans << '\n';
    for (int x = 0; x < (1 << 13); ++x) {
      if (dp[x] != 0x3f3f3f3f) {
        cout << x << ' ';
      }
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int tests = 1;
    for (int tc = 1; tc <= tests; ++tc) {
      TestCase();
    }
    return 0;
  }
  