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>
  #pragma GCC optimize(3)
  using namespace std;
  inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
  inline int qpow(int a, int n) {
    int ans = 1;
    while (n) {
      if (n & 1) {
        ans *= a;
      }
      a *= a;
      n >>= 1;
    }
    return ans;
  }
  inline long long _qpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  inline long long ksm(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  const int maxnum = 1e6 + 5;
  const int MAXN = (1 << 13) - 1;
  set<int> st[maxnum];
  int n;
  int x;
  int f[MAXN + 5];
  void slove() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> x;
      st[x].insert(i);
    }
    for (int i = 0; i <= MAXN; i++) f[i] = 0x3f3f3f3f;
    f[0] = 0;
    for (int i = 0; i <= MAXN; i++) {
      for (int j = MAXN; j >= 0; j--) {
        auto it = st[i].upper_bound(f[j]);
        if (it != st[i].end()) {
          f[i ^ j] = min(f[i ^ j], *it);
        }
      }
    }
    vector<int> ans;
    for (int i = 0; i <= MAXN; i++) {
      if (f[i] != 0x3f3f3f3f) ans.push_back(i);
    }
    cout << ans.size() << "\n";
    for (int x : ans) cout << x << " ";
    cout << "\n";
  }
  signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    ;
    slove();
    return 0;
  }
  