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 <class c>
  struct rge {
    c b, e;
  };
  template <class c>
  rge<c> range(c i, c j) {
    return rge<c>{i, j};
  }
  template <class c>
  auto dud(c* x) -> decltype(cerr << *x, 0);
  template <class c>
  char dud(...);
  struct debug {
    template <class c>
    debug& operator<<(const c&) {
      return *this;
    }
  };
  long long power(long long x, long long y, long long M) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) res = (res * x) % M;
      y >>= 1;
      x = (x * x) % M;
    }
    return res % M;
  }
  long long power(long long x, long long y) {
    long long res = 1;
    while (y > 0) {
      if (y & 1) res = (res * x);
      y >>= 1;
      x = (x * x);
    }
    return res;
  }
  const long long mxn = 1e6 + 6, mod = 1e9 + 7, inf = 1e18;
  long long t, casee = 1;
  void test_case() {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long mx = 0;
    for (long long i = 0; i <= 5000; i++) {
      for (long long j = 0; j < 5000; j++) {
        mx = max(mx, (i ^ j) + 10);
      }
    }
    debug() << " ["
            << "mx"
               ": "
            << (mx) << "] ";
    vector<vector<long long>> possiblesubsetxor(mx, vector<long long>());
    for (long long i = 0; i < mx; i++) possiblesubsetxor[i].push_back(0);
    vector<long long> pos(mx, 0);
    vector<long long> help(mx, mx);
    pos[0] = 1;
    help[0] = 0;
    for (long long i = 0; i < n; i++) {
      for (long long currxor : possiblesubsetxor[a[i]]) {
        long long newxor = currxor ^ a[i];
        pos[newxor] = 1;
        if (a[i] < help[newxor]) {
          for (long long k = a[i] + 1; k < help[newxor]; k++) {
            possiblesubsetxor[k].push_back(newxor);
          }
          help[newxor] = a[i];
        }
      }
      possiblesubsetxor[a[i]].clear();
    }
    long long cnt = 0;
    for (long long i = 0; i < mx; i++) cnt += pos[i];
    cout << cnt << "\n";
    for (long long i = 0; i < mx; i++) {
      if (pos[i]) cout << i << " ";
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    t = 1;
    while (t--) {
      test_case();
      casee++;
    }
    return 0;
  }
  