#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e6 + 50, M = 8200;
int n, a[N];
vector<int> xs[M];
vector<pair<int, int> > ear;
bool vis[M];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < (n); ++i) cin >> a[i], xs[a[i]].push_back(i);
  ear.push_back({0, -1});
  for (int v = 0; v < (M); ++v) {
    fill(vis, vis + M, 0);
    vector<pair<int, int> > gear;
    int c = 0;
    for (int i : xs[v]) {
      int r = c;
      while (r < (int)(ear).size() && ear[r].second < i) {
        if (!vis[ear[r].first]) {
          gear.push_back({ear[r]});
          vis[ear[r].first] = true;
        }
        r++;
      }
      for (int j = c; j < r; j++) {
        int nxt = ear[j].first ^ v;
        if (!vis[nxt]) {
          gear.push_back({nxt, i});
          vis[nxt] = true;
        }
      }
      c = r;
    }
    while (c < (int)(ear).size()) {
      if (!vis[ear[c].first]) {
        gear.push_back({ear[c]});
        vis[ear[c].first] = true;
      }
      c++;
    }
    ear = gear;
  }
  sort(begin(ear), end(ear));
  cout << (int)(ear).size() << '\n';
  for (int i = 0; i < ((int)(ear).size()); ++i) cout << ear[i].first << " ";
  cout << endl;
}

### Prompt

Please formulate a cpp solution 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 N = (int)1e6 + 50, M = 8200;
int n, a[N];
vector<int> xs[M];
vector<pair<int, int> > ear;
bool vis[M];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < (n); ++i) cin >> a[i], xs[a[i]].push_back(i);
  ear.push_back({0, -1});
  for (int v = 0; v < (M); ++v) {
    fill(vis, vis + M, 0);
    vector<pair<int, int> > gear;
    int c = 0;
    for (int i : xs[v]) {
      int r = c;
      while (r < (int)(ear).size() && ear[r].second < i) {
        if (!vis[ear[r].first]) {
          gear.push_back({ear[r]});
          vis[ear[r].first] = true;
        }
        r++;
      }
      for (int j = c; j < r; j++) {
        int nxt = ear[j].first ^ v;
        if (!vis[nxt]) {
          gear.push_back({nxt, i});
          vis[nxt] = true;
        }
      }
      c = r;
    }
    while (c < (int)(ear).size()) {
      if (!vis[ear[c].first]) {
        gear.push_back({ear[c]});
        vis[ear[c].first] = true;
      }
      c++;
    }
    ear = gear;
  }
  sort(begin(ear), end(ear));
  cout << (int)(ear).size() << '\n';
  for (int i = 0; i < ((int)(ear).size()); ++i) cout << ear[i].first << " ";
  cout << endl;
}

```