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;
  template <typename T>
  inline void read(T &x) {
    x = 0;
    char c = getchar();
    bool f = false;
    for (; !isdigit(c); c = getchar()) f |= c == '-';
    for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    if (f) x = -x;
  }
  template <typename T, typename... Arg>
  inline void read(T &x, Arg &...args) {
    read(x);
    read(args...);
  }
  template <typename T>
  inline void write(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  const int N = 8200;
  int n, a[1001000];
  vector<int> p[N];
  int dp[2][N];
  signed main() {
    read(n);
    for (int i = 1; i <= n; i++) read(a[i]), p[a[i]].push_back(i);
    int now = 0, lst = 1;
    memset(dp, 0x3f, sizeof dp);
    dp[now][0] = 0;
    for (int i = 0; i < N; i++) {
      if (!p[i].size()) continue;
      swap(now, lst);
      memcpy(dp[now], dp[lst], sizeof dp[now]);
      for (int j = 0; j < N; j++) {
        if (p[i].back() < dp[lst][j]) continue;
        ((dp[now][j ^ i]) =
             min((dp[now][j ^ i]),
                 (*lower_bound(p[i].begin(), p[i].end(), dp[lst][j]))));
      }
    }
    vector<int> ans;
    for (int i = 0; i < N; i++)
      if (dp[now][i] != 0x3f3f3f3f) ans.push_back(i);
    write(ans.size());
    putchar('\n');
    for (auto x : ans) write(x), putchar(' ');
  }
  