#include <bits/stdc++.h>
using namespace std;
const int maxn = 8192;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<vector<int>> p(maxn);
  for (int i = 0; i < n; ++i) cin >> a[i], p[a[i]].push_back(i);
  vector<int> dp(maxn, n);
  dp[0] = -1;
  for (int i = 1; i <= 5000; ++i) {
    vector<int> nw = dp;
    for (int j = 0; j < maxn; ++j)
      if (dp[j] <= n) {
        auto nx = lower_bound(p[i].begin(), p[i].end(), dp[j]);
        if (nx != p[i].end()) {
          nw[i ^ j] = min(nw[i ^ j], *nx);
        }
      }
    swap(nw, dp);
  }
  vector<int> ans;
  for (int i = 0; i < maxn; ++i)
    if (dp[i] < n) ans.push_back(i);
  cout << ans.size() << '\n';
  for (int i : ans) cout << i << " ";
  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;
const int maxn = 8192;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<vector<int>> p(maxn);
  for (int i = 0; i < n; ++i) cin >> a[i], p[a[i]].push_back(i);
  vector<int> dp(maxn, n);
  dp[0] = -1;
  for (int i = 1; i <= 5000; ++i) {
    vector<int> nw = dp;
    for (int j = 0; j < maxn; ++j)
      if (dp[j] <= n) {
        auto nx = lower_bound(p[i].begin(), p[i].end(), dp[j]);
        if (nx != p[i].end()) {
          nw[i ^ j] = min(nw[i ^ j], *nx);
        }
      }
    swap(nw, dp);
  }
  vector<int> ans;
  for (int i = 0; i < maxn; ++i)
    if (dp[i] < n) ans.push_back(i);
  cout << ans.size() << '\n';
  for (int i : ans) cout << i << " ";
  return 0;
}

```