#include <bits/stdc++.h>
using namespace std;
const int MAXV = 5001, MAXXOR = (1 << 13);
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> A(n + 1);
  vector<int> values;
  values.reserve(MAXV);
  vector<vector<int>> pos(MAXV);
  for (int i = 1; i <= n; ++i) {
    cin >> A[i];
    if (pos[A[i]].empty()) values.push_back(A[i]);
    pos[A[i]].push_back(i);
  }
  sort(values.begin(), values.end());
  vector<int> smallest_idx(MAXXOR, n + 1);
  smallest_idx[0] = 0;
  for (const int &x : values) {
    for (int cur = 0; cur < MAXXOR; ++cur) {
      int id = smallest_idx[cur ^ x];
      auto new_id = upper_bound(pos[x].begin(), pos[x].end(), id);
      if (new_id != pos[x].end()) {
        smallest_idx[cur] = min(smallest_idx[cur], *new_id);
      }
    }
  }
  vector<int> res;
  res.reserve(MAXXOR);
  for (int i = 0; i < MAXXOR; ++i)
    if (smallest_idx[i] != n + 1) res.push_back(i);
  cout << res.size() << '\n';
  for (const int &x : res) cout << x << ' ';
  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 MAXV = 5001, MAXXOR = (1 << 13);
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> A(n + 1);
  vector<int> values;
  values.reserve(MAXV);
  vector<vector<int>> pos(MAXV);
  for (int i = 1; i <= n; ++i) {
    cin >> A[i];
    if (pos[A[i]].empty()) values.push_back(A[i]);
    pos[A[i]].push_back(i);
  }
  sort(values.begin(), values.end());
  vector<int> smallest_idx(MAXXOR, n + 1);
  smallest_idx[0] = 0;
  for (const int &x : values) {
    for (int cur = 0; cur < MAXXOR; ++cur) {
      int id = smallest_idx[cur ^ x];
      auto new_id = upper_bound(pos[x].begin(), pos[x].end(), id);
      if (new_id != pos[x].end()) {
        smallest_idx[cur] = min(smallest_idx[cur], *new_id);
      }
    }
  }
  vector<int> res;
  res.reserve(MAXXOR);
  for (int i = 0; i < MAXXOR; ++i)
    if (smallest_idx[i] != n + 1) res.push_back(i);
  cout << res.size() << '\n';
  for (const int &x : res) cout << x << ' ';
  return 0;
}

```