#include <bits/stdc++.h>
using namespace std;
int n, a[1000005], A = 1, m;
bool in[5005][1 << 13], s[1 << 13];
vector<int> g[5005];
int main() {
  scanf("%d", &n);
  s[0] = 1;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), m = max(a[i], m);
  for (int i = 0; i <= m; i++) g[i].push_back(0), in[i][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = (int)g[a[i]].size() - 1; ~j; j--) {
      int v = a[i] ^ g[a[i]][j];
      if (!s[v]) s[v] = 1, A++;
      for (int k = a[i] + 1; k <= m; k++) {
        if (in[k][v]) break;
        in[k][v] = 1;
        g[k].push_back(v);
      }
      g[a[i]].pop_back();
    }
  }
  printf("%d\n", A);
  for (int i = 0; i < (1 << 13); i++)
    if (s[i]) printf("%d ", i);
  puts("");
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int n, a[1000005], A = 1, m;
bool in[5005][1 << 13], s[1 << 13];
vector<int> g[5005];
int main() {
  scanf("%d", &n);
  s[0] = 1;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), m = max(a[i], m);
  for (int i = 0; i <= m; i++) g[i].push_back(0), in[i][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = (int)g[a[i]].size() - 1; ~j; j--) {
      int v = a[i] ^ g[a[i]][j];
      if (!s[v]) s[v] = 1, A++;
      for (int k = a[i] + 1; k <= m; k++) {
        if (in[k][v]) break;
        in[k][v] = 1;
        g[k].push_back(v);
      }
      g[a[i]].pop_back();
    }
  }
  printf("%d\n", A);
  for (int i = 0; i < (1 << 13); i++)
    if (s[i]) printf("%d ", i);
  puts("");
  return 0;
}

```