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 N = 1e6 + 100, M = 1 << 13;
  int n, a[N], MIN[M], f[M];
  vector<int> p[M];
  inline void ckmin(int &a, int b) { a = ((a < b) ? a : b); }
  inline void ckmax(int &a, int b) { a = ((a > b) ? a : b); }
  inline int read() {
    int f = 1, x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s >= '0' && s <= '9') {
      x = x * 10 + s - '0';
      s = getchar();
    }
    return x * f;
  }
  signed main() {
    n = read();
    for (int i = 1; i <= n; i++) a[i] = read(), p[a[i]].push_back(i);
    memset(MIN, 0x3f, sizeof(MIN));
    MIN[0] = 0;
    for (int i = 1; i <= 5000; i++)
      if (!p[i].empty()) {
        vector<int> b = p[i];
        memset(f, 0x3f, sizeof(f));
        for (int j = 0; j < M; j++)
          if (MIN[j] != 0x3f3f3f3f) {
            int pos = lower_bound(b.begin(), b.end(), MIN[j]) - b.begin();
            if (pos == (int)b.size())
              f[j ^ i] = 0x3f3f3f3f;
            else
              f[j ^ i] = b[pos];
          }
        for (int j = 0; j < M; j++) ckmin(MIN[j], f[j]);
      }
    vector<int> ans;
    for (int i = 0; i < M; i++)
      if (MIN[i] != 0x3f3f3f3f) ans.push_back(i);
    printf("%d\n", (int)ans.size());
    for (int i : ans) printf("%d ", i);
    printf("\n");
  }
  