#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using ll = long long;
using ld = long double;
using namespace std;
const ll inf = int(2e9) + 1;
int main() {
  ll n;
  cin >> n;
  vector<ll> s;
  map<int, vector<ll>> dd;
  for (int i = 0; i < n; i++) {
    int q;
    cin >> q;
    dd[q].push_back(i);
  }
  set<pair<int, int>> q;
  for (int i = 1; i <= 5000; i++) {
    for (int j = 0; j < 5; j++) {
      if (dd[i].size() == 0) continue;
      q.insert(make_pair(dd[i][(ll)round((dd[i].size() - 1) * j / 4)], i));
    }
  }
  for (auto el : q) s.push_back(el.second);
  ;
  n = s.size();
  vector<ll> used(10130, -inf);
  for (int i = n - 1; i >= 0; i--) {
    used[s[i]] = max(used[s[i]], s[i]);
    for (int j = 1; j <= 10120; j++) {
      if (used[j] > s[i]) {
        used[j ^ s[i]] = max(used[j ^ s[i]], s[i]);
      }
    }
  }
  ll res = 0;
  used[0] = 1;
  for (int i = 0; i <= 10120; i++)
    if (used[i] != -inf) res++;
  cout << res << endl;
  for (int i = 0; i <= 10120; i++)
    if (used[i] != -inf) cout << i << ' ';
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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] 

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using ll = long long;
using ld = long double;
using namespace std;
const ll inf = int(2e9) + 1;
int main() {
  ll n;
  cin >> n;
  vector<ll> s;
  map<int, vector<ll>> dd;
  for (int i = 0; i < n; i++) {
    int q;
    cin >> q;
    dd[q].push_back(i);
  }
  set<pair<int, int>> q;
  for (int i = 1; i <= 5000; i++) {
    for (int j = 0; j < 5; j++) {
      if (dd[i].size() == 0) continue;
      q.insert(make_pair(dd[i][(ll)round((dd[i].size() - 1) * j / 4)], i));
    }
  }
  for (auto el : q) s.push_back(el.second);
  ;
  n = s.size();
  vector<ll> used(10130, -inf);
  for (int i = n - 1; i >= 0; i--) {
    used[s[i]] = max(used[s[i]], s[i]);
    for (int j = 1; j <= 10120; j++) {
      if (used[j] > s[i]) {
        used[j ^ s[i]] = max(used[j ^ s[i]], s[i]);
      }
    }
  }
  ll res = 0;
  used[0] = 1;
  for (int i = 0; i <= 10120; i++)
    if (used[i] != -inf) res++;
  cout << res << endl;
  for (int i = 0; i <= 10120; i++)
    if (used[i] != -inf) cout << i << ' ';
  return 0;
}

```