#include <bits/stdc++.h>
using namespace std;
const int oo = 1e9;
const int N = 5001;
const int maxxor = (1 << 13);
int f[2][maxxor];
vector<int> a[N];
int cnp(int l, int r, int pos, int id) {
  if (l > r) return oo;
  int mid = (l + r) / 2;
  if (a[id][mid] > pos) return min(a[id][mid], cnp(l, mid - 1, pos, id));
  return cnp(mid + 1, r, pos, id);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    a[x].push_back(i);
  }
  for (int i = 0; i <= 1; i++)
    for (int j = 0; j < maxxor; j++) f[i][j] = oo;
  f[0][0] = 0;
  int id;
  vector<int> res;
  for (int i = 1; i < N; i++) {
    int now = i & 1;
    int back = now ^ 1;
    for (int j = 0; j < maxxor; j++) {
      f[now][j] = f[back][j];
      if (a[i].size() == 0) {
        if (i == N - 1 && f[now][j] != oo) res.push_back(j);
        continue;
      }
      int pos = f[back][j ^ i];
      id = cnp(0, a[i].size() - 1, pos, i);
      if (id != oo) f[now][j] = min(f[now][j], id);
      if (i == N - 1 && f[now][j] != oo) res.push_back(j);
    }
  }
  cout << res.size() << endl;
  for (int xnew : res) cout << xnew << " ";
  return 0;
}

### Prompt

Your challenge is to write 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;
const int oo = 1e9;
const int N = 5001;
const int maxxor = (1 << 13);
int f[2][maxxor];
vector<int> a[N];
int cnp(int l, int r, int pos, int id) {
  if (l > r) return oo;
  int mid = (l + r) / 2;
  if (a[id][mid] > pos) return min(a[id][mid], cnp(l, mid - 1, pos, id));
  return cnp(mid + 1, r, pos, id);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    a[x].push_back(i);
  }
  for (int i = 0; i <= 1; i++)
    for (int j = 0; j < maxxor; j++) f[i][j] = oo;
  f[0][0] = 0;
  int id;
  vector<int> res;
  for (int i = 1; i < N; i++) {
    int now = i & 1;
    int back = now ^ 1;
    for (int j = 0; j < maxxor; j++) {
      f[now][j] = f[back][j];
      if (a[i].size() == 0) {
        if (i == N - 1 && f[now][j] != oo) res.push_back(j);
        continue;
      }
      int pos = f[back][j ^ i];
      id = cnp(0, a[i].size() - 1, pos, i);
      if (id != oo) f[now][j] = min(f[now][j], id);
      if (i == N - 1 && f[now][j] != oo) res.push_back(j);
    }
  }
  cout << res.size() << endl;
  for (int xnew : res) cout << xnew << " ";
  return 0;
}

```