#include <bits/stdc++.h>
using namespace std;
const int maxa = 1 << 13;
vector<vector<int> > v(maxa);
int x[maxa], x2[maxa];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i], v[a[i]].push_back(i);
  fill(x, x + maxa, n);
  x[0] = -1;
  for (int i = 0; i < maxa; i++)
    if (v.size()) {
      fill(x2, x2 + maxa, n);
      for (int j = 0; j < maxa; j++)
        if (x[j] != n) {
          int l = upper_bound(v[i].begin(), v[i].end(), x[j]) - v[i].begin();
          if (l != v[i].size()) x2[j ^ i] = v[i][l];
        }
      for (int j = 0; j < maxa; j++) x[j] = min(x2[j], x[j]);
    }
  vector<int> b;
  for (int i = 0; i < maxa; i++)
    if (x[i] < n) b.push_back(i);
  cout << b.size() << "\n";
  for (int i : b) cout << i << " ";
  cout << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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 maxa = 1 << 13;
vector<vector<int> > v(maxa);
int x[maxa], x2[maxa];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i], v[a[i]].push_back(i);
  fill(x, x + maxa, n);
  x[0] = -1;
  for (int i = 0; i < maxa; i++)
    if (v.size()) {
      fill(x2, x2 + maxa, n);
      for (int j = 0; j < maxa; j++)
        if (x[j] != n) {
          int l = upper_bound(v[i].begin(), v[i].end(), x[j]) - v[i].begin();
          if (l != v[i].size()) x2[j ^ i] = v[i][l];
        }
      for (int j = 0; j < maxa; j++) x[j] = min(x2[j], x[j]);
    }
  vector<int> b;
  for (int i = 0; i < maxa; i++)
    if (x[i] < n) b.push_back(i);
  cout << b.size() << "\n";
  for (int i : b) cout << i << " ";
  cout << "\n";
  return 0;
}

```