#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    int o;
    cin >> o;
    a[i] = o;
  }
  int mx = 8191;
  vector<bool> can[mx + 1];
  vector<bool> was[mx + 1];
  for (int j = 0; j < mx + 1; ++j) {
    can[j].resize(mx + 1);
    was[j].resize(mx + 1);
  }
  vector<int> cc[mx + 1];
  cc[0].push_back(0);
  for (int m = 0; m < n; ++m) {
    cc[a[m]].push_back(0);
    can[a[m]][0] = true;
  }
  for (int k = 0; k < n; ++k) {
    for (int i = 0; i < cc[a[k]].size(); ++i) {
      int f = cc[a[k]][i] ^ a[k];
      for (int j = a[k] + 1; j < mx + 1; ++j) {
        if (can[j][f] || was[j][f]) break;
        cc[j].push_back(f);
        can[j][f] = true;
      }
      was[a[k]][cc[a[k]][i]] = true;
    }
    cc[a[k]].clear();
  }
  vector<bool> used(mx + 1);
  for (int k1 = 0; k1 < mx + 1; ++k1) {
    for (int l = 0; l < mx + 1; ++l) {
      if (can[k1][l]) used[l] = true;
    }
  }
  int ccc = 0;
  for (int l1 = 0; l1 < mx + 1; ++l1) {
    ccc += used[l1];
  }
  cout << ccc << "\n";
  for (int l1 = 0; l1 < mx + 1; ++l1) {
    if (used[l1]) {
      cout << l1 << " ";
    }
  }
  cout << "\n";
}

### Prompt

In CPP, your task is to solve 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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    int o;
    cin >> o;
    a[i] = o;
  }
  int mx = 8191;
  vector<bool> can[mx + 1];
  vector<bool> was[mx + 1];
  for (int j = 0; j < mx + 1; ++j) {
    can[j].resize(mx + 1);
    was[j].resize(mx + 1);
  }
  vector<int> cc[mx + 1];
  cc[0].push_back(0);
  for (int m = 0; m < n; ++m) {
    cc[a[m]].push_back(0);
    can[a[m]][0] = true;
  }
  for (int k = 0; k < n; ++k) {
    for (int i = 0; i < cc[a[k]].size(); ++i) {
      int f = cc[a[k]][i] ^ a[k];
      for (int j = a[k] + 1; j < mx + 1; ++j) {
        if (can[j][f] || was[j][f]) break;
        cc[j].push_back(f);
        can[j][f] = true;
      }
      was[a[k]][cc[a[k]][i]] = true;
    }
    cc[a[k]].clear();
  }
  vector<bool> used(mx + 1);
  for (int k1 = 0; k1 < mx + 1; ++k1) {
    for (int l = 0; l < mx + 1; ++l) {
      if (can[k1][l]) used[l] = true;
    }
  }
  int ccc = 0;
  for (int l1 = 0; l1 < mx + 1; ++l1) {
    ccc += used[l1];
  }
  cout << ccc << "\n";
  for (int l1 = 0; l1 < mx + 1; ++l1) {
    if (used[l1]) {
      cout << l1 << " ";
    }
  }
  cout << "\n";
}

```