#include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
const int N = 1e6 + 100, K = 10001, INF = 2e18, MOD = 1e9 + 7;
int n, a[N];
vector<bool> ans(K);
vector<vector<bool>> ok(K, vector<bool>(K));
vector<int> v[K];
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < K; i++) {
    ok[i][0] = true;
    v[i].push_back(0);
  }
  for (int i = 0; i < n; i++) {
    for (int x : v[a[i]]) {
      int e = (x ^ a[i]);
      for (int h = a[i] + 1; h < K; h++) {
        if (ok[h][e]) break;
        if (!ok[h][e]) {
          ok[h][e] = true;
          v[h].push_back(e);
        }
      }
    }
    v[a[i]].clear();
  }
  for (int i = 0; i < K; i++) {
    for (int j = 0; j < K; j++) {
      if (ok[i][j]) {
        ans[j] = true;
      }
    }
  }
  int cnt = 0;
  for (int i = 0; i < K; i++) {
    cnt += ans[i];
  }
  cout << cnt << '\n';
  for (int i = 0; i < K; i++) {
    if (ans[i]) {
      cout << i << " ";
    }
  }
}

### 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;
void solve();
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
const int N = 1e6 + 100, K = 10001, INF = 2e18, MOD = 1e9 + 7;
int n, a[N];
vector<bool> ans(K);
vector<vector<bool>> ok(K, vector<bool>(K));
vector<int> v[K];
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < K; i++) {
    ok[i][0] = true;
    v[i].push_back(0);
  }
  for (int i = 0; i < n; i++) {
    for (int x : v[a[i]]) {
      int e = (x ^ a[i]);
      for (int h = a[i] + 1; h < K; h++) {
        if (ok[h][e]) break;
        if (!ok[h][e]) {
          ok[h][e] = true;
          v[h].push_back(e);
        }
      }
    }
    v[a[i]].clear();
  }
  for (int i = 0; i < K; i++) {
    for (int j = 0; j < K; j++) {
      if (ok[i][j]) {
        ans[j] = true;
      }
    }
  }
  int cnt = 0;
  for (int i = 0; i < K; i++) {
    cnt += ans[i];
  }
  cout << cnt << '\n';
  for (int i = 0; i < K; i++) {
    if (ans[i]) {
      cout << i << " ";
    }
  }
}

```