#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int N = (2 << 13);
int can[N];
vector<int> q[N];
int mn[N];
void Set(int p, int j) {
  if (p >= mn[j]) return;
  for (int tp = p; tp < mn[j]; tp++) {
    if (tp + 1 < N && mn[j ^ (tp + 1)] > tp + 1) {
      q[tp + 1].push_back(j ^ (tp + 1));
    }
  }
  mn[j] = p;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  fill(mn, mn + N, N);
  int n;
  cin >> n;
  Set(0, 0);
  while (n--) {
    int p;
    cin >> p;
    for (auto j : q[p]) {
      Set(p, j);
    }
    q[p].clear();
  }
  int cnt = 0;
  for (int j = 0; j < N; j++) {
    can[j] = (mn[j] != N);
    cnt += can[j];
  }
  cout << cnt << '\n';
  for (int j = 0; j < N; j++) {
    if (can[j]) {
      cout << j << ' ';
    }
  }
  cout << '\n';
}

### Prompt

Generate a CPP solution 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>
#pragma GCC optimize("Ofast")
using namespace std;
const int N = (2 << 13);
int can[N];
vector<int> q[N];
int mn[N];
void Set(int p, int j) {
  if (p >= mn[j]) return;
  for (int tp = p; tp < mn[j]; tp++) {
    if (tp + 1 < N && mn[j ^ (tp + 1)] > tp + 1) {
      q[tp + 1].push_back(j ^ (tp + 1));
    }
  }
  mn[j] = p;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  fill(mn, mn + N, N);
  int n;
  cin >> n;
  Set(0, 0);
  while (n--) {
    int p;
    cin >> p;
    for (auto j : q[p]) {
      Set(p, j);
    }
    q[p].clear();
  }
  int cnt = 0;
  for (int j = 0; j < N; j++) {
    can[j] = (mn[j] != N);
    cnt += can[j];
  }
  cout << cnt << '\n';
  for (int j = 0; j < N; j++) {
    if (can[j]) {
      cout << j << ' ';
    }
  }
  cout << '\n';
}

```