#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long INF = 1e18 + 123;
const long double EPS = 1e-9;
const int inf = 1e9 + 123;
const int MOD = 1e9 + 7;
const int N = 1e5 + 123;
const int M = 1e6 + 1;
const int K = 8192;
const int Q = 5001;
const double pi = 3.14159265359;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
bool was[K][Q], ok[K];
vector<int> v[N];
void solve() {
  int n;
  cin >> n;
  ok[0] = 1;
  for (int second = 1; second < Q; second++) {
    v[second].push_back(0);
    was[0][second] = 1;
  }
  for (int i = 1, a; i <= n; i++) {
    cin >> a;
    for (int first : v[a]) {
      ok[first ^ a] = 1;
      for (int second = a + 1; second < Q && !was[first ^ a][second];
           second++) {
        was[first ^ a][second] = 1;
        v[second].push_back(first ^ a);
      }
    }
    v[a].clear();
  }
  vector<int> v;
  for (int first = 0; first < K; first++) {
    if (ok[first]) {
      v.push_back(first);
    }
  }
  cout << (int)v.size() << '\n';
  for (int first : v) cout << first << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long INF = 1e18 + 123;
const long double EPS = 1e-9;
const int inf = 1e9 + 123;
const int MOD = 1e9 + 7;
const int N = 1e5 + 123;
const int M = 1e6 + 1;
const int K = 8192;
const int Q = 5001;
const double pi = 3.14159265359;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
bool was[K][Q], ok[K];
vector<int> v[N];
void solve() {
  int n;
  cin >> n;
  ok[0] = 1;
  for (int second = 1; second < Q; second++) {
    v[second].push_back(0);
    was[0][second] = 1;
  }
  for (int i = 1, a; i <= n; i++) {
    cin >> a;
    for (int first : v[a]) {
      ok[first ^ a] = 1;
      for (int second = a + 1; second < Q && !was[first ^ a][second];
           second++) {
        was[first ^ a][second] = 1;
        v[second].push_back(first ^ a);
      }
    }
    v[a].clear();
  }
  vector<int> v;
  for (int first = 0; first < K; first++) {
    if (ok[first]) {
      v.push_back(first);
    }
  }
  cout << (int)v.size() << '\n';
  for (int first : v) cout << first << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```