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 double pi = acos(-1.0);
  const int inf = 0x3f3f3f3f;
  const long long INF = 0x3f3f3f3f3f3f3f3fll;
  int dir[4][2] = {{1, 0}, {0, -1}, {-1, 0}, {0, 1}};
  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 x * f;
  }
  void wri(long long x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) wri(x / 10);
    putchar(x % 10 + '0');
  }
  void write(long long x) {
    wri(x);
    puts("");
  }
  long long ksc(long long a, long long b, long long mod) {
    return ((a * b -
             (long long)((long long)((long double)a / mod * b + 1e-3) * mod)) %
                mod +
            mod) %
           mod;
  }
  long long ksm1(long long a, long long n, long long mod) {
    long long t = 1;
    for (; n; n >>= 1, (a *= a) %= mod)
      if (n & 1) (t *= a) %= mod;
    return t;
  }
  long long ksm2(long long a, long long b, long long mod) {
    long long ints = 1;
    while (b) {
      if (b & 1) ints = ksc(ints, a, mod);
      b >>= 1, a = ksc(a, a, mod);
    }
    return ints % mod;
  }
  int is2pow(long long x) { return x > 0 ? (x & (x - 1)) == 0 : 0; }
  long long lowbit(long long x) { return x & -x; }
  long long myrand(long long l, long long r) {
    static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
    return uniform_int_distribution<long long>(l, r)(gen);
  }
  const int N = 1e6 + 5;
  const int M = 8192;
  pair<int, int> arr[N];
  int num[M];
  int brr[N];
  int n;
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> arr[i].first, arr[i].second = i;
    sort(arr + 1, arr + 1 + n);
    memset(num, inf, sizeof(num));
    int pos = 1;
    num[0] = 0;
    while (pos <= n) {
      int tmp = 1;
      int mark = arr[pos].first;
      brr[tmp] = arr[pos].second;
      pos++;
      while (pos <= n && arr[pos].first == mark)
        brr[++tmp] = arr[pos].second, pos++;
      for (int i = 0; i < 8192; i++)
        if (num[i] < brr[tmp])
          num[i ^ mark] =
              min(num[i ^ mark],
                  brr[lower_bound(brr + 1, brr + 1 + tmp, num[i]) - brr]);
    }
    int ans = 0;
    for (int i = 0; i < 8192; i++)
      if (num[i] != inf) ans++;
    cout << ans << endl;
    for (int i = 0; i < 8192; i++)
      if (num[i] != inf) cout << i << " ";
    cout << endl;
    return 0;
  }
  