#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n;
int a[N];
int f[10000];
vector<int> nw[5005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < 8192; ++i) f[i] = 5001;
  f[0] = 0;
  for (int i = 1; i <= 5000; ++i) nw[i].push_back(0);
  for (int i = 1; i <= n; ++i) {
    int sz = nw[a[i]].size();
    for (int j = 0; j < sz; ++j) {
      int tmp = nw[a[i]][j] ^ a[i];
      for (int k = a[i] + 1; k <= f[tmp]; ++k) {
        nw[k].push_back(tmp);
      }
      f[tmp] = min(f[tmp], a[i]);
    }
    nw[a[i]].clear();
  }
  int ans = 0;
  for (int i = 0; i < 8192; ++i)
    if (f[i] != 5001) ans++;
  printf("%d\n", ans);
  for (int i = 0; i < 8192; ++i)
    if (f[i] != 5001) printf("%d ", i);
  puts("");
  return 0;
}

### 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;
const int N = 1e6 + 5;
int n;
int a[N];
int f[10000];
vector<int> nw[5005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < 8192; ++i) f[i] = 5001;
  f[0] = 0;
  for (int i = 1; i <= 5000; ++i) nw[i].push_back(0);
  for (int i = 1; i <= n; ++i) {
    int sz = nw[a[i]].size();
    for (int j = 0; j < sz; ++j) {
      int tmp = nw[a[i]][j] ^ a[i];
      for (int k = a[i] + 1; k <= f[tmp]; ++k) {
        nw[k].push_back(tmp);
      }
      f[tmp] = min(f[tmp], a[i]);
    }
    nw[a[i]].clear();
  }
  int ans = 0;
  for (int i = 0; i < 8192; ++i)
    if (f[i] != 5001) ans++;
  printf("%d\n", ans);
  for (int i = 0; i < 8192; ++i)
    if (f[i] != 5001) printf("%d ", i);
  puts("");
  return 0;
}

```