#include <bits/stdc++.h>
using namespace std;
int n;
int a[1000001];
queue<int> q[8192];
int f[8192];
void update(int x, int y) {
  if (f[x] > 5000) f[x] = 5000;
  while (f[x] > y) {
    q[f[x]].push(x);
    f[x]--;
  }
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < 8192; i++) f[i] = 5001;
  update(0, 0);
  for (int i = 1, old; i <= n; i++) {
    while (!q[a[i]].empty()) {
      int x = q[a[i]].front();
      q[a[i]].pop();
      update(x ^ a[i], a[i]);
    }
  }
  vector<int> ans;
  for (int i = 0; i < 8192; i++)
    if (f[i] <= 5000) ans.push_back(i);
  cout << ans.size() << '\n';
  for (auto&& x : ans) cout << x << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Create a solution in cpp for 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 n;
int a[1000001];
queue<int> q[8192];
int f[8192];
void update(int x, int y) {
  if (f[x] > 5000) f[x] = 5000;
  while (f[x] > y) {
    q[f[x]].push(x);
    f[x]--;
  }
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < 8192; i++) f[i] = 5001;
  update(0, 0);
  for (int i = 1, old; i <= n; i++) {
    while (!q[a[i]].empty()) {
      int x = q[a[i]].front();
      q[a[i]].pop();
      update(x ^ a[i], a[i]);
    }
  }
  vector<int> ans;
  for (int i = 0; i < 8192; i++)
    if (f[i] <= 5000) ans.push_back(i);
  cout << ans.size() << '\n';
  for (auto&& x : ans) cout << x << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

```