#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> pool(8200, 0);
  pool[0] = 1;
  vector<vector<int>> has(n);
  vector<vector<int>> nxt(5005);
  vector<int> minop(8200, 1e9);
  for (int i = n - 1; i >= 0; i--) {
    nxt[a[i]].emplace_back(i);
  }
  for (int i = 1; i <= 5000; i++) {
    if (nxt[i].size() != 0) {
      has[nxt[i].back()].emplace_back(i);
    }
  }
  for (int i = 0; i < n; i++) {
    nxt[a[i]].pop_back();
    for (auto x : has[i]) {
      if (minop[x] <= a[i] + 1) continue;
      pool[x]++;
      for (int j = a[i] + 1; j <= min(minop[x], 5000); j++) {
        if (nxt[j].size() != 0) {
          has[nxt[j].back()].emplace_back(j ^ x);
        }
      }
      minop[x] = a[i] + 1;
    }
  }
  vector<int> sol;
  for (int i = 0; i < 8200; i++)
    if (pool[i]) sol.emplace_back(i);
  cout << sol.size() << '\n';
  for (auto x : sol) cout << x << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) solve();
}

### Prompt

Please provide a Cpp coded solution 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;
using ll = long long;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> pool(8200, 0);
  pool[0] = 1;
  vector<vector<int>> has(n);
  vector<vector<int>> nxt(5005);
  vector<int> minop(8200, 1e9);
  for (int i = n - 1; i >= 0; i--) {
    nxt[a[i]].emplace_back(i);
  }
  for (int i = 1; i <= 5000; i++) {
    if (nxt[i].size() != 0) {
      has[nxt[i].back()].emplace_back(i);
    }
  }
  for (int i = 0; i < n; i++) {
    nxt[a[i]].pop_back();
    for (auto x : has[i]) {
      if (minop[x] <= a[i] + 1) continue;
      pool[x]++;
      for (int j = a[i] + 1; j <= min(minop[x], 5000); j++) {
        if (nxt[j].size() != 0) {
          has[nxt[j].back()].emplace_back(j ^ x);
        }
      }
      minop[x] = a[i] + 1;
    }
  }
  vector<int> sol;
  for (int i = 0; i < 8200; i++)
    if (pool[i]) sol.emplace_back(i);
  cout << sol.size() << '\n';
  for (auto x : sol) cout << x << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) solve();
}

```