#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const int lim = 1e5;
int n, x, dr[(1 << 14)], ans[(1 << 14)];
vector<int> v[(1 << 14)];
int main() {
  cin >> n;
  for (int i = 0; i < (1 << 13); i++) v[i].push_back(0);
  for (int i = 0; i < (1 << 13); i++) dr[i] = (1 << 13);
  dr[0] = 0;
  ans[0] = 1;
  int j;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    for (auto it : v[x]) {
      if (dr[it ^ x] < x) continue;
      for (j = x + 1; j <= dr[it ^ x]; ++j) v[j].push_back(it ^ x);
      ans[it ^ x] = 1;
      dr[it ^ x] = x;
    }
    v[x].clear();
  }
  int cnt = 0;
  for (int i = 0; i < (1 << 13); i++)
    if (ans[i]) cnt++;
  cout << cnt << "\n";
  for (int i = 0; i < (1 << 13); i++)
    if (ans[i]) cout << i << " ";
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long inf = 1e18;
const int lim = 1e5;
int n, x, dr[(1 << 14)], ans[(1 << 14)];
vector<int> v[(1 << 14)];
int main() {
  cin >> n;
  for (int i = 0; i < (1 << 13); i++) v[i].push_back(0);
  for (int i = 0; i < (1 << 13); i++) dr[i] = (1 << 13);
  dr[0] = 0;
  ans[0] = 1;
  int j;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    for (auto it : v[x]) {
      if (dr[it ^ x] < x) continue;
      for (j = x + 1; j <= dr[it ^ x]; ++j) v[j].push_back(it ^ x);
      ans[it ^ x] = 1;
      dr[it ^ x] = x;
    }
    v[x].clear();
  }
  int cnt = 0;
  for (int i = 0; i < (1 << 13); i++)
    if (ans[i]) cnt++;
  cout << cnt << "\n";
  for (int i = 0; i < (1 << 13); i++)
    if (ans[i]) cout << i << " ";
  return 0;
}

```