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;
  long long mod = pow(10, 9) + 7;
  int gcdExtended(int a, int b, int *x, int *y);
  int modInverse(int b, int m) {
    int x, y;
    int g = gcdExtended(b, m, &x, &y);
    if (g != 1) return -1;
    return (x % m + m) % m;
  }
  int gcdExtended(int a, int b, int *x, int *y) {
    if (a == 0) {
      *x = 0, *y = 1;
      return b;
    }
    int x1, y1;
    int gcd = gcdExtended(b % a, a, &x1, &y1);
    *x = y1 - (b / a) * x1;
    *y = x1;
    return gcd;
  }
  long long sumbincoef(int N, int k) {
    long long bincoef = 1, sum = 1;
    int i;
    for (i = 1; i < k; i++) {
      long long val = ((N - i + 1) % mod * modInverse(i, mod)) % mod;
      bincoef = (bincoef % mod * val) % mod;
      if (i % 2 == 0) sum = (sum + bincoef) % mod;
    }
    return sum % mod;
  }
  const int N = 300000 + 10;
  long long factorialNumInverse[N + 1];
  long long naturalNumInverse[N + 1];
  long long fact[N + 1];
  void InverseofNumber(long long p) {
    naturalNumInverse[0] = naturalNumInverse[1] = 1;
    for (int i = 2; i <= N; i++)
      naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p / i) % p;
  }
  void InverseofFactorial(long long p) {
    factorialNumInverse[0] = factorialNumInverse[1] = 1;
    for (int i = 2; i <= N; i++)
      factorialNumInverse[i] =
          (naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
  }
  void factorial(long long p) {
    fact[0] = 1;
    for (int i = 1; i <= N; i++) {
      fact[i] = (fact[i - 1] * i) % p;
    }
  }
  long long Binomial(long long N, long long R, long long p) {
    long long ans =
        ((fact[N] * factorialNumInverse[R]) % p * factorialNumInverse[N - R]) % p;
    return ans;
  }
  int parent[N], ran[N];
  void make_set(int v) {
    parent[v] = v;
    ran[v] = 0;
  }
  int find_set(int v) {
    if (v == parent[v]) return v;
    return parent[v] = find_set(parent[v]);
  }
  void union_sets(int a, int b) {
    a = find_set(a);
    b = find_set(b);
    if (a != b) {
      if (ran[a] < ran[b]) swap(a, b);
      parent[b] = a;
      if (ran[a] == ran[b]) ran[a]++;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    mt19937 rng(
        (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
    long long p = mod;
    InverseofNumber(p);
    InverseofFactorial(p);
    factorial(p);
    int t = 1;
    while (t > 0) {
      int n;
      cin >> n;
      vector<int> a(n);
      vector<int> dp(8192, 1e6);
      unordered_map<int, vector<int>> mapa;
      for (int i = 0; i < n; i++) {
        cin >> a[i];
        mapa[a[i]].push_back(i);
      }
      dp[0] = -1;
      for (long long i = 0; i <= 5000; i++) {
        for (long long j = 0; j < 8192; j++) {
          auto it = upper_bound(mapa[i].begin(), mapa[i].end(), dp[j]);
          if (it == mapa[i].end()) continue;
          dp[i ^ j] = min(dp[i ^ j], *it);
        }
      }
      vector<int> ans;
      for (int i = 0; i < 8192; i++) {
        if (dp[i] < 1e6) {
          ans.push_back(i);
        }
      }
      cout << ans.size() << "\n";
      for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " ";
      }
      cout << "\n";
      t--;
    }
  }
  