#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e+16;
const int SMALL_INF = 1998244353;
const int NSIZE = 4 * 1e+5 + 1000;
const long long MOD = 998244353;
const long double EPS = 1e-12;
const int MAX = 1e+6 + 1000;
const long double PI = atan(1) * 4;
const long long HZ = -1692281337;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> xpos(10001, SMALL_INF);
  xpos[0] = -1;
  vector<vector<int> > apos(5001);
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    apos[a].push_back(i);
  }
  for (int i = 0; i <= 5000; i++) {
    if (apos[i].empty()) continue;
    for (int j = 0; j <= 10000; j++) {
      if ((i ^ j) <= 10000 && xpos[i ^ j] < apos[i].back()) {
        xpos[j] = min(
            xpos[j], *upper_bound(apos[i].begin(), apos[i].end(), xpos[i ^ j]));
      }
    }
  }
  vector<int> ans;
  for (int i = 0; i <= 10000; i++) {
    if (xpos[i] != SMALL_INF) ans.push_back(i);
  }
  cout << ans.size() << '\n';
  for (auto i : ans) cout << i << " ";
}

### 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;
const long long INF = 1e+16;
const int SMALL_INF = 1998244353;
const int NSIZE = 4 * 1e+5 + 1000;
const long long MOD = 998244353;
const long double EPS = 1e-12;
const int MAX = 1e+6 + 1000;
const long double PI = atan(1) * 4;
const long long HZ = -1692281337;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> xpos(10001, SMALL_INF);
  xpos[0] = -1;
  vector<vector<int> > apos(5001);
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    apos[a].push_back(i);
  }
  for (int i = 0; i <= 5000; i++) {
    if (apos[i].empty()) continue;
    for (int j = 0; j <= 10000; j++) {
      if ((i ^ j) <= 10000 && xpos[i ^ j] < apos[i].back()) {
        xpos[j] = min(
            xpos[j], *upper_bound(apos[i].begin(), apos[i].end(), xpos[i ^ j]));
      }
    }
  }
  vector<int> ans;
  for (int i = 0; i <= 10000; i++) {
    if (xpos[i] != SMALL_INF) ans.push_back(i);
  }
  cout << ans.size() << '\n';
  for (auto i : ans) cout << i << " ";
}

```