#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 500;
int n;
unordered_map<int, int> ans;
vector<int> buc[5050];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1, tmp; i <= n; i++) {
    cin >> tmp;
    buc[tmp].push_back(i);
  }
  ans.clear();
  ans[0] = 0;
  for (int i = 0; i <= 5000; i++) {
    if (buc[i].empty()) continue;
    vector<pair<int, int> > newitem;
    newitem.clear();
    for (auto &[x, y] : ans) {
      auto it = upper_bound(buc[i].begin(), buc[i].end(), y);
      if (it == buc[i].end()) continue;
      newitem.push_back({x ^ i, *it});
    }
    for (auto &it : newitem) {
      if (ans.find(it.first) == ans.end()) {
        ans[it.first] = it.second;
      } else {
        ans[it.first] = min(ans[it.first], it.second);
      }
    }
  }
  vector<int> res;
  for (auto &[key, val] : ans) res.push_back(key);
  cout << res.size() << '\n';
  sort(res.begin(), res.end());
  for (auto &it : res) cout << it << " ";
  cout << '\n';
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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>
using namespace std;
const int maxn = 1e6 + 500;
int n;
unordered_map<int, int> ans;
vector<int> buc[5050];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1, tmp; i <= n; i++) {
    cin >> tmp;
    buc[tmp].push_back(i);
  }
  ans.clear();
  ans[0] = 0;
  for (int i = 0; i <= 5000; i++) {
    if (buc[i].empty()) continue;
    vector<pair<int, int> > newitem;
    newitem.clear();
    for (auto &[x, y] : ans) {
      auto it = upper_bound(buc[i].begin(), buc[i].end(), y);
      if (it == buc[i].end()) continue;
      newitem.push_back({x ^ i, *it});
    }
    for (auto &it : newitem) {
      if (ans.find(it.first) == ans.end()) {
        ans[it.first] = it.second;
      } else {
        ans[it.first] = min(ans[it.first], it.second);
      }
    }
  }
  vector<int> res;
  for (auto &[key, val] : ans) res.push_back(key);
  cout << res.size() << '\n';
  sort(res.begin(), res.end());
  for (auto &it : res) cout << it << " ";
  cout << '\n';
  return 0;
}

```