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>
  long long read() {
    long long x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') x = x * 10 + (c - '0'), c = getchar();
    return f * x;
  }
  long long min(long long a, long long b) { return a < b ? a : b; }
  long long max(long long a, long long b) { return a > b ? a : b; }
  template <typename T>
  bool umin(T& a, T t) {
    if (a > t) return a = t, 1;
    return 0;
  }
  template <typename T>
  bool umax(T& a, T t) {
    if (a < t) return a = t, 1;
    return 0;
  }
  const int MAXN = 5011, B = 85;
  typedef std::bitset<(1 << 13)> bits;
  bits a[MAXN], nxt[MAXN], res;
  std::vector<short> seq[MAXN];
  int main() {
    int n = read();
    for (int i = 0; i <= 5000; ++i) nxt[i][0] = 1, seq[i].emplace_back(0);
    a[0][0] = 1;
    while (n--) {
      int x = read();
      for (auto y : seq[x]) {
        int k = y ^ x;
        a[x][k] = 1;
        for (int i = x + 1; i <= 5000; ++i)
          if (!nxt[i][k])
            nxt[i][k] = 1, seq[i].emplace_back(k);
          else
            break;
      }
      seq[x].clear();
    }
    for (int i = 0; i <= 5000; ++i) res |= a[i];
    printf("%d\n", int(res.count()));
    for (int i = 0; i < (1 << 13); ++i)
      if (res[i]) printf("%d ", i);
    puts("");
    return 0;
  }
  