#include <bits/stdc++.h>
using namespace std;
using li = long long;
using ld = long double;
using pii = pair<int, int>;
const int INF = 1e9 + 13;
const li INF64 = 1e18 + 13;
const int N = 1e5 + 13;
const int K = 500;
const int A = 9000;
vector<int> wh[A];
int mn[A];
int mn1[A];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    wh[a[i]].push_back(i);
  }
  mn[0] = -1;
  for (int i = 1; i < A; i++) {
    mn[i] = n;
  }
  for (int x = 0; x < A; x++) {
    for (int y = 0; y < A; y++) mn1[y] = mn[y];
    for (int y = 0; y < A; y++) {
      int i = upper_bound(wh[x].begin(), wh[x].end(), mn[y]) - wh[x].begin();
      if (i < wh[x].size()) mn1[y ^ x] = min(mn[y ^ x], wh[x][i]);
    }
    for (int y = 0; y < A; y++) mn[y] = mn1[y];
  }
  int res = 0;
  for (int i = 0; i < A; i++) res += (mn[i] < n);
  cout << res << endl;
  for (int i = 0; i < A; i++) {
    if (mn[i] < n) cout << i << ' ';
  }
  cout << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 li = long long;
using ld = long double;
using pii = pair<int, int>;
const int INF = 1e9 + 13;
const li INF64 = 1e18 + 13;
const int N = 1e5 + 13;
const int K = 500;
const int A = 9000;
vector<int> wh[A];
int mn[A];
int mn1[A];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    wh[a[i]].push_back(i);
  }
  mn[0] = -1;
  for (int i = 1; i < A; i++) {
    mn[i] = n;
  }
  for (int x = 0; x < A; x++) {
    for (int y = 0; y < A; y++) mn1[y] = mn[y];
    for (int y = 0; y < A; y++) {
      int i = upper_bound(wh[x].begin(), wh[x].end(), mn[y]) - wh[x].begin();
      if (i < wh[x].size()) mn1[y ^ x] = min(mn[y ^ x], wh[x][i]);
    }
    for (int y = 0; y < A; y++) mn[y] = mn1[y];
  }
  int res = 0;
  for (int i = 0; i < A; i++) res += (mn[i] < n);
  cout << res << endl;
  for (int i = 0; i < A; i++) {
    if (mn[i] < n) cout << i << ' ';
  }
  cout << endl;
}

```