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 canhazfast() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  }
  template <typename T>
  T gcd(T a, T b) {
    return b ? gcd(b, a % b) : a;
  }
  template <typename T>
  T extgcd(T a, T b, T &x, T &y) {
    T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
    while (b) {
      T q = a / b;
      a %= b;
      swap(a, b);
      x0 -= q * x1;
      swap(x0, x1);
      y0 -= q * y1;
      swap(y0, y1);
    }
    x = x0;
    y = y0;
    return a;
  }
  int ctz(unsigned x) { return __builtin_ctz(x); }
  int ctzll(unsigned long long x) { return __builtin_ctzll(x); }
  int clz(unsigned x) { return __builtin_clz(x); }
  int clzll(unsigned long long x) { return __builtin_clzll(x); }
  int popcnt(unsigned x) { return __builtin_popcount(x); }
  int popcntll(unsigned long long x) { return __builtin_popcountll(x); }
  int bsr(unsigned x) { return 31 ^ clz(x); }
  int bsrll(unsigned long long x) { return 63 ^ clzll(x); }
  vector<int> q[5000 + 1];
  bool res[8192];
  int ub[8192];
  int main() {
    canhazfast();
    int n;
    vector<int> ans;
    cin >> n;
    for (int i = 1; i <= 5000; ++i) q[i].push_back(0);
    res[0] = true;
    fill_n(ub, 8192, 5000);
    for (int i = 0; i < n; ++i) {
      int a;
      cin >> a;
      if (!a) continue;
      for (int v : q[a]) {
        int x = v ^ a;
        if (x == 0) continue;
        res[x] = true;
        for (int j = a + 1; j <= ub[x]; ++j) q[j].push_back(x);
        ub[x] = min(ub[x], a);
      }
      q[a].clear();
    }
    for (int i = 0; i < 8192; ++i) {
      if (res[i]) ans.push_back(i);
    }
    cout << ans.size() << '\n';
    for (int x : ans) cout << x << ' ';
    cout << '\n';
    return 0;
  }
  