#include <bits/stdc++.h>
using namespace std;
inline long long md(long long m, long long v) { return ((v % m) + m) % m; }
const int MAX = 8200;
const int NOT = 1e9;
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  vector<vector<int> > idx(MAX + 1);
  for (int i = 0; i < n; i++) {
    idx[arr[i]].push_back(i);
  }
  vector<int> dp(MAX + 1, NOT);
  dp[0] = -1;
  for (int a = 1; a <= MAX; a++) {
    int I = a % 2, J = 1 - I;
    if ((int)(idx[a]).size() == 0) continue;
    int upper = idx[a].back();
    dp[a] = min(dp[a], idx[a].front());
    for (int b = 1; b <= MAX; b++) {
      int target = a ^ b;
      if (target > MAX) continue;
      if (dp[target] < upper) {
        int where =
            int(lower_bound((idx[a]).begin(), (idx[a]).end(), dp[target]) -
                begin(idx[a]));
        dp[b] = min(dp[b], idx[a][where]);
      }
    }
  }
  vector<int> ans;
  for (int i = 0; i <= MAX; i++) {
    if (dp[i] != NOT) ans.push_back(i);
  }
  cout << (int)(ans).size() << "\n";
  for (int i : ans) cout << i << ' ';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  solve();
  return 0;
}

### Prompt

Please formulate 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;
inline long long md(long long m, long long v) { return ((v % m) + m) % m; }
const int MAX = 8200;
const int NOT = 1e9;
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  vector<vector<int> > idx(MAX + 1);
  for (int i = 0; i < n; i++) {
    idx[arr[i]].push_back(i);
  }
  vector<int> dp(MAX + 1, NOT);
  dp[0] = -1;
  for (int a = 1; a <= MAX; a++) {
    int I = a % 2, J = 1 - I;
    if ((int)(idx[a]).size() == 0) continue;
    int upper = idx[a].back();
    dp[a] = min(dp[a], idx[a].front());
    for (int b = 1; b <= MAX; b++) {
      int target = a ^ b;
      if (target > MAX) continue;
      if (dp[target] < upper) {
        int where =
            int(lower_bound((idx[a]).begin(), (idx[a]).end(), dp[target]) -
                begin(idx[a]));
        dp[b] = min(dp[b], idx[a][where]);
      }
    }
  }
  vector<int> ans;
  for (int i = 0; i <= MAX; i++) {
    if (dp[i] != NOT) ans.push_back(i);
  }
  cout << (int)(ans).size() << "\n";
  for (int i : ans) cout << i << ' ';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  solve();
  return 0;
}

```