#include <bits/stdc++.h>
using namespace std;
const int mx = 8192;
int n, dp[10000], q[10000], p[10000], a[1000010], res;
vector<int> v[5010];
bool vis[10000];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    v[a[i]].push_back(i);
  }
  dp[0] = 0;
  for (int i = 1; i <= mx; i++) dp[i] = n + 1, q[i] = i;
  for (int i = 1; i <= 5000; i++) {
    int tot = 0, now = -1;
    memset(vis, 0, sizeof(vis));
    for (int j = 0; j < v[i].size(); j++) {
      int x = v[i][j], lst = --tot;
      while (tot <= mx && dp[q[++tot]] < x)
        if (!vis[q[tot]]) p[++now] = q[tot];
      for (int j = lst; j < tot; j++) {
        if (dp[q[j] ^ i] > x) {
          p[++now] = q[j] ^ i;
          dp[p[now]] = x;
          vis[p[now]] = 1;
        }
      }
    }
    --tot;
    while (++tot <= mx)
      if (!vis[q[tot]]) p[++now] = q[tot];
    for (int i = 0; i <= mx; i++) q[i] = p[i];
  }
  for (int i = 0; i <= mx; i++)
    if (dp[i] <= n) res++;
  cout << res << endl;
  for (int i = 0; i <= mx; i++)
    if (dp[i] <= n) printf("%d ", i);
}

### 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;
const int mx = 8192;
int n, dp[10000], q[10000], p[10000], a[1000010], res;
vector<int> v[5010];
bool vis[10000];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    v[a[i]].push_back(i);
  }
  dp[0] = 0;
  for (int i = 1; i <= mx; i++) dp[i] = n + 1, q[i] = i;
  for (int i = 1; i <= 5000; i++) {
    int tot = 0, now = -1;
    memset(vis, 0, sizeof(vis));
    for (int j = 0; j < v[i].size(); j++) {
      int x = v[i][j], lst = --tot;
      while (tot <= mx && dp[q[++tot]] < x)
        if (!vis[q[tot]]) p[++now] = q[tot];
      for (int j = lst; j < tot; j++) {
        if (dp[q[j] ^ i] > x) {
          p[++now] = q[j] ^ i;
          dp[p[now]] = x;
          vis[p[now]] = 1;
        }
      }
    }
    --tot;
    while (++tot <= mx)
      if (!vis[q[tot]]) p[++now] = q[tot];
    for (int i = 0; i <= mx; i++) q[i] = p[i];
  }
  for (int i = 0; i <= mx; i++)
    if (dp[i] <= n) res++;
  cout << res << endl;
  for (int i = 0; i <= mx; i++)
    if (dp[i] <= n) printf("%d ", i);
}

```