#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e6 + 1;
const int B = 8191;
int n, a[MAXN];
vector<pair<int, int> > q[MAXN];
int eachq[B + 1];
queue<int> app[5005];
int dp[5005][B + 1];
bool vis[B + 1];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), app[a[i]].push(i);
  for (int i = 0; i <= 5000; ++i) app[i].push(INF);
  for (int i = 0; i <= 5000; ++i)
    for (int j = 0; j <= B; ++j) eachq[i ^ j] = i;
  memset(dp, 63, sizeof(dp));
  int cnt = 0;
  for (int i = 0; i <= 5000; ++i) {
    dp[i][i] = app[i].front();
    if (dp[i][i] != INF) {
      q[dp[i][i]].push_back(make_pair(i, i));
      cnt += 1;
      vis[i] = 1;
    }
  }
  for (int t = 1; t <= n; ++t) {
    for (auto it : q[t]) {
      while (eachq[it.second] != -1 && eachq[it.second] > it.first) {
        int Tmp_ = eachq[it.second]--;
        auto Tmp = make_pair(Tmp_, it.second ^ Tmp_);
        while (app[Tmp.first].front() <= t) app[Tmp.first].pop();
        if (app[Tmp.first].front() != INF) {
          cnt += vis[Tmp.second] == 0;
          vis[Tmp.second] = 1;
          if (cnt == B + 1) break;
          dp[Tmp.first][Tmp.second] = app[Tmp.first].front();
          q[dp[Tmp.first][Tmp.second]].push_back(Tmp);
        }
      }
    }
  }
  vector<int> ans;
  ans.push_back(0);
  for (int i = 1; i <= B; ++i) {
    if (vis[i]) ans.push_back(i);
  }
  printf("%d\n", int(ans.size()));
  for (auto it : ans) printf("%d ", it);
  return 0;
}

### Prompt

Please create a solution in cpp to 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 INF = 0x3f3f3f3f;
const int MAXN = 1e6 + 1;
const int B = 8191;
int n, a[MAXN];
vector<pair<int, int> > q[MAXN];
int eachq[B + 1];
queue<int> app[5005];
int dp[5005][B + 1];
bool vis[B + 1];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), app[a[i]].push(i);
  for (int i = 0; i <= 5000; ++i) app[i].push(INF);
  for (int i = 0; i <= 5000; ++i)
    for (int j = 0; j <= B; ++j) eachq[i ^ j] = i;
  memset(dp, 63, sizeof(dp));
  int cnt = 0;
  for (int i = 0; i <= 5000; ++i) {
    dp[i][i] = app[i].front();
    if (dp[i][i] != INF) {
      q[dp[i][i]].push_back(make_pair(i, i));
      cnt += 1;
      vis[i] = 1;
    }
  }
  for (int t = 1; t <= n; ++t) {
    for (auto it : q[t]) {
      while (eachq[it.second] != -1 && eachq[it.second] > it.first) {
        int Tmp_ = eachq[it.second]--;
        auto Tmp = make_pair(Tmp_, it.second ^ Tmp_);
        while (app[Tmp.first].front() <= t) app[Tmp.first].pop();
        if (app[Tmp.first].front() != INF) {
          cnt += vis[Tmp.second] == 0;
          vis[Tmp.second] = 1;
          if (cnt == B + 1) break;
          dp[Tmp.first][Tmp.second] = app[Tmp.first].front();
          q[dp[Tmp.first][Tmp.second]].push_back(Tmp);
        }
      }
    }
  }
  vector<int> ans;
  ans.push_back(0);
  for (int i = 1; i <= B; ++i) {
    if (vis[i]) ans.push_back(i);
  }
  printf("%d\n", int(ans.size()));
  for (auto it : ans) printf("%d ", it);
  return 0;
}

```