#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pi = pair<ll, ll>;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll minf = 0xc0c0c0c0c0c0c0c0;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ll n;
  cin >> n;
  vector<vector<ll>> v(5050);
  vector<ll> dp(10010, inf);
  for (ll i = 1; i <= n; i++) {
    ll a;
    cin >> a;
    v[a].push_back(i);
  }
  dp[0] = -1;
  for (ll i = 0; i <= 5000; i++) {
    for (ll j = 0; j <= 8192; j++) {
      auto it = upper_bound((v[i]).begin(), (v[i]).end(), dp[j]);
      if (it != v[i].end()) {
        dp[i ^ j] = min(dp[i ^ j], *it);
      }
    }
  }
  vector<ll> res;
  for (ll i = 0; i <= 8192; i++) {
    if (dp[i] != inf) {
      res.push_back(i);
    }
  }
  cout << ll((res).size()) << '\n';
  for (auto i : res) {
    cout << i << ' ';
  }
  cout << '\n';
  return 0;
}

### 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;
using ll = long long;
using pi = pair<ll, ll>;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll minf = 0xc0c0c0c0c0c0c0c0;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ll n;
  cin >> n;
  vector<vector<ll>> v(5050);
  vector<ll> dp(10010, inf);
  for (ll i = 1; i <= n; i++) {
    ll a;
    cin >> a;
    v[a].push_back(i);
  }
  dp[0] = -1;
  for (ll i = 0; i <= 5000; i++) {
    for (ll j = 0; j <= 8192; j++) {
      auto it = upper_bound((v[i]).begin(), (v[i]).end(), dp[j]);
      if (it != v[i].end()) {
        dp[i ^ j] = min(dp[i ^ j], *it);
      }
    }
  }
  vector<ll> res;
  for (ll i = 0; i <= 8192; i++) {
    if (dp[i] != inf) {
      res.push_back(i);
    }
  }
  cout << ll((res).size()) << '\n';
  for (auto i : res) {
    cout << i << ' ';
  }
  cout << '\n';
  return 0;
}

```