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;
  inline int rd() {
    int x = 0;
    char ch, t = 0;
    while (!isdigit(ch = getchar())) t |= ch == '-';
    while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
    return x = t ? -x : x;
  }
  inline long long maxll(long long x, long long y) { return x > y ? x : y; }
  inline long long minll(long long x, long long y) { return x < y ? x : y; }
  inline long long absll(long long x) { return x > 0ll ? x : -x; }
  inline long long gcd(long long x, long long y) {
    return (y == 0) ? x : gcd(y, x % y);
  }
  int n, All = 8191;
  int a[1000005];
  bitset<8200> ans[8200], vis[8200];
  vector<int> pre[1000005];
  int main() {
    n = rd();
    for (int i = 0; i <= All; i++) pre[i].push_back(0);
    for (int i = 1; i <= n; i++) {
      a[i] = rd();
      if (!vis[a[i]][a[i]]) pre[a[i]].push_back(a[i]), vis[a[i]][a[i]] = true;
    }
    for (int i = 1; i <= n; i++) {
      for (int j : pre[a[i]]) {
        ans[a[i]][j] = true;
        for (int k = a[i] + 1; k <= 5000; k++) {
          if (vis[k][k ^ j]) break;
          vis[k][k ^ j] = true, pre[k].push_back(k ^ j);
        }
      }
      pre[a[i]].clear();
    }
    for (int i = 1; i <= 5000; i++) ans[0] |= ans[i];
    printf("%lld\n", ans[0].count());
    for (int i = 0; i <= All; i++)
      if (ans[0][i]) printf("%d ", i);
    printf("\n");
    return 0;
  }
  