#include <bits/stdc++.h>
using namespace std;
const int MAX = 8192;
vector<int> upd[MAX];
vector<int> least(MAX, MAX);
vector<bool> ok(MAX, false);
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int z = 0; z < MAX; z++) {
    upd[z].push_back(0);
  }
  least[0] = 0;
  ok[0] = true;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    for (int x : upd[v[i]]) {
      int p = x ^ v[i];
      ok[p] = true;
      for (int z = v[i] + 1; z < least[p]; z++) upd[z].push_back(p);
      least[p] = min(v[i], least[p]);
    }
    upd[v[i]].clear();
  }
  {
    int cnt = 0;
    for (int z = 0; z < MAX; z++) cnt += ok[z];
    cout << cnt << '\n';
    for (int z = 0; z < MAX; z++) {
      if (ok[z]) cout << z << ' ';
    }
    cout << '\n';
  }
  return;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  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>
using namespace std;
const int MAX = 8192;
vector<int> upd[MAX];
vector<int> least(MAX, MAX);
vector<bool> ok(MAX, false);
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int z = 0; z < MAX; z++) {
    upd[z].push_back(0);
  }
  least[0] = 0;
  ok[0] = true;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    for (int x : upd[v[i]]) {
      int p = x ^ v[i];
      ok[p] = true;
      for (int z = v[i] + 1; z < least[p]; z++) upd[z].push_back(p);
      least[p] = min(v[i], least[p]);
    }
    upd[v[i]].clear();
  }
  {
    int cnt = 0;
    for (int z = 0; z < MAX; z++) cnt += ok[z];
    cout << cnt << '\n';
    for (int z = 0; z < MAX; z++) {
      if (ok[z]) cout << z << ' ';
    }
    cout << '\n';
  }
  return;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```