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 INF = 1e9;
  const int N = 1e6 + 7;
  const int MOD = 1e9 + 7;
  const int LOG = 16;
  const int maxn = (1 << 16);
  inline long long int Max(const long long int &x, const long long int &y) {
    return x > y ? x : y;
  }
  inline long long int Min(const long long int &x, const long long int &y) {
    return x > y ? y : x;
  }
  inline int read() {
    char ch = getchar();
    int x = 0, f = 1;
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while ('0' <= ch && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  inline long long int readl() {
    char ch = getchar();
    long long int x = 0, f = 1;
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while ('0' <= ch && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  int A[N], mn[maxn];
  vector<int> res, pos[N];
  set<int> second;
  int find(int i, int a) {
    int l = 0, r = pos[i].size() - 1;
    if (a >= pos[i][r]) return r + 1;
    if (a < pos[i][0]) return 0;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (pos[i][mid] == a) {
        return mid + 1;
      }
      if (pos[i][mid] < a && pos[i][mid + 1] > a) {
        return mid + 1;
      }
      if (pos[i][mid] < a) {
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
    return r;
  }
  int main() {
    int n = read();
    for (int i = 0; i < maxn; i++) {
      mn[i] = INF;
    }
    for (int i = 1; i <= n; i++) {
      A[i] = read();
      pos[A[i]].push_back(i);
    }
    for (int i = 0; i <= 5000; i++) {
      if (pos[i].size() == 0) continue;
      mn[i] = min(mn[i], pos[i][0]);
      for (int j = 0; j < maxn; j++) {
        if (mn[j] != INF) {
          int _pos = find(i, mn[j]);
          if (_pos < 0 || _pos > pos[i].size() - 1) continue;
          _pos = pos[i][_pos];
          if (_pos > mn[j]) {
            int tmp = (i ^ j);
            mn[tmp] = min(mn[tmp], _pos);
          }
        }
      }
    }
    res.push_back(0);
    for (int i = 1; i < maxn; i++) {
      if (mn[i] != INF) {
        res.push_back(i);
      }
    }
    cout << res.size() << endl;
    for (auto x : res) {
      cout << x << ' ';
    }
    cout << endl;
    return 0;
  }
  