#include <bits/stdc++.h>
using namespace std;
using ull = unsigned long long;
using ll = long long;
using pii = pair<int, int>;
const int MXINF = 0x3f3f3f3f;
const int MNINF = 0x7f7f7f7f;
const ull MOD = 1000000007;
const int MAXN = 8192;
vector<int> g[MAXN + 5];
int ans[MAXN + 5], r[MAXN + 5];
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i <= MAXN; i++) {
    ans[i] = 0;
    r[i] = MAXN;
    g[i].push_back(0);
  }
  ans[0] = 1;
  for (int i = 0; i < n; i++) {
    for (auto x : g[a[i]]) {
      int val = x ^ a[i];
      ans[val] = 1;
      while (r[val] > a[i]) {
        g[r[val]].push_back(val);
        r[val]--;
      }
    }
    g[a[i]] = {};
  }
  int cnt = 0;
  for (int i = 0; i <= MAXN; i++) cnt += ans[i];
  cout << cnt << endl;
  for (int i = 0; i <= MAXN; i++)
    if (ans[i]) cout << i << " ";
  cout << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
using ull = unsigned long long;
using ll = long long;
using pii = pair<int, int>;
const int MXINF = 0x3f3f3f3f;
const int MNINF = 0x7f7f7f7f;
const ull MOD = 1000000007;
const int MAXN = 8192;
vector<int> g[MAXN + 5];
int ans[MAXN + 5], r[MAXN + 5];
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i <= MAXN; i++) {
    ans[i] = 0;
    r[i] = MAXN;
    g[i].push_back(0);
  }
  ans[0] = 1;
  for (int i = 0; i < n; i++) {
    for (auto x : g[a[i]]) {
      int val = x ^ a[i];
      ans[val] = 1;
      while (r[val] > a[i]) {
        g[r[val]].push_back(val);
        r[val]--;
      }
    }
    g[a[i]] = {};
  }
  int cnt = 0;
  for (int i = 0; i <= MAXN; i++) cnt += ans[i];
  cout << cnt << endl;
  for (int i = 0; i <= MAXN; i++)
    if (ans[i]) cout << i << " ";
  cout << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}

```