#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
int f[5005][8205], n;
vector<int> g[5005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    g[x].push_back(i);
  }
  for (int i = 0; i <= 5000; i++)
    for (int j = 1; j <= 8200; j++) f[i][j] = inf;
  int t = 0;
  for (int i = 1; i <= 5000; i++) {
    if (!g[i].size()) continue;
    for (int j = 1; j <= 8191; j++) {
      int yu = j ^ i, kk = f[t][yu];
      if (t) f[i][j] = f[t][j];
      if ((kk != inf) && (t || !yu) && (yu <= 8191)) {
        int l = 0, r = g[i].size() - 1;
        while (l <= r) {
          int mid = (l + r) / 2;
          if (g[i][mid] <= kk)
            l = mid + 1;
          else
            r = mid - 1;
        }
        if (l < g[i].size()) f[i][j] = min(f[i][j], g[i][l]);
      }
    }
    t = i;
  }
  int ans = 0;
  for (int i = 0; i <= 8191; i++)
    if (f[t][i] != inf) ans++;
  cout << ans << endl;
  for (int i = 0; i <= 8191; i++)
    if (f[t][i] != inf) cout << 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 inf = 1e9 + 7;
int f[5005][8205], n;
vector<int> g[5005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    g[x].push_back(i);
  }
  for (int i = 0; i <= 5000; i++)
    for (int j = 1; j <= 8200; j++) f[i][j] = inf;
  int t = 0;
  for (int i = 1; i <= 5000; i++) {
    if (!g[i].size()) continue;
    for (int j = 1; j <= 8191; j++) {
      int yu = j ^ i, kk = f[t][yu];
      if (t) f[i][j] = f[t][j];
      if ((kk != inf) && (t || !yu) && (yu <= 8191)) {
        int l = 0, r = g[i].size() - 1;
        while (l <= r) {
          int mid = (l + r) / 2;
          if (g[i][mid] <= kk)
            l = mid + 1;
          else
            r = mid - 1;
        }
        if (l < g[i].size()) f[i][j] = min(f[i][j], g[i][l]);
      }
    }
    t = i;
  }
  int ans = 0;
  for (int i = 0; i <= 8191; i++)
    if (f[t][i] != inf) ans++;
  cout << ans << endl;
  for (int i = 0; i <= 8191; i++)
    if (f[t][i] != inf) cout << i << ' ';
  puts("");
  return 0;
}

```