#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<int> A(n);
  vector<vector<int>> idx(5001);
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    idx[A[i]].push_back(i);
  }
  for (int i = 0; i <= 5000; ++i) {
    sort(idx[i].begin(), idx[i].end());
  }
  vector<int> dp(8192, -1), dpn;
  dp[0] = 1e9;
  dpn = dp;
  for (int v = 5000; v >= 0; --v) {
    if (idx[v].empty()) continue;
    for (int t = 0; t < 8192; ++t) {
      int id = dp[t ^ v];
      if (id == -1) continue;
      auto it = lower_bound(idx[v].begin(), idx[v].end(), id);
      if (it == idx[v].begin()) continue;
      it = prev(it);
      dpn[t] = max(*it, dp[t]);
    }
    dp = dpn;
  }
  int cnt = 0;
  for (int t = 0; t < 8192; ++t)
    if (dp[t] != -1) cnt++;
  cout << cnt << endl;
  for (int t = 0; t < 8192; ++t)
    if (dp[t] != -1) cout << t << " ";
  cout << endl;
}

### Prompt

Your task is to create 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;
int main() {
  int n;
  cin >> n;
  vector<int> A(n);
  vector<vector<int>> idx(5001);
  for (int i = 0; i < n; i++) {
    cin >> A[i];
    idx[A[i]].push_back(i);
  }
  for (int i = 0; i <= 5000; ++i) {
    sort(idx[i].begin(), idx[i].end());
  }
  vector<int> dp(8192, -1), dpn;
  dp[0] = 1e9;
  dpn = dp;
  for (int v = 5000; v >= 0; --v) {
    if (idx[v].empty()) continue;
    for (int t = 0; t < 8192; ++t) {
      int id = dp[t ^ v];
      if (id == -1) continue;
      auto it = lower_bound(idx[v].begin(), idx[v].end(), id);
      if (it == idx[v].begin()) continue;
      it = prev(it);
      dpn[t] = max(*it, dp[t]);
    }
    dp = dpn;
  }
  int cnt = 0;
  for (int t = 0; t < 8192; ++t)
    if (dp[t] != -1) cnt++;
  cout << cnt << endl;
  for (int t = 0; t < 8192; ++t)
    if (dp[t] != -1) cout << t << " ";
  cout << endl;
}

```