#include <bits/stdc++.h>
using namespace std;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
constexpr int MAXN = 1e6 + 5;
constexpr int MOD = 1e9 + 7;
constexpr int INF = 1e9 + 1;
constexpr int LOG = 3e1 + 1;
constexpr int SQ = 500;
constexpr int ALP = 5e3 + 3;
int n;
int a[MAXN];
bool mark[ALP][ALP * 2], has[ALP * 2];
vector<int> G[ALP], ans;
void add(int idx, int x) {
  for (int i = idx + 1; i <= 5000; i++) {
    if (!mark[i][x]) {
      G[i].push_back(x);
      mark[i][x] = true;
    } else {
      break;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= 5000; i++) {
    G[i].push_back(0);
    mark[i][0] = true;
  }
  ans.push_back(0);
  has[0] = true;
  for (int i = 1; i <= n; i++) {
    int ptr = 0;
    while (ptr < G[a[i]].size()) {
      int x = G[a[i]][ptr++] ^ a[i];
      if (!has[x]) {
        ans.push_back(x);
        has[x] = true;
      }
      add(a[i], x);
    }
    G[a[i]].clear();
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << "\n";
  for (auto x : ans) {
    cout << x << " ";
  }
  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;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
constexpr int MAXN = 1e6 + 5;
constexpr int MOD = 1e9 + 7;
constexpr int INF = 1e9 + 1;
constexpr int LOG = 3e1 + 1;
constexpr int SQ = 500;
constexpr int ALP = 5e3 + 3;
int n;
int a[MAXN];
bool mark[ALP][ALP * 2], has[ALP * 2];
vector<int> G[ALP], ans;
void add(int idx, int x) {
  for (int i = idx + 1; i <= 5000; i++) {
    if (!mark[i][x]) {
      G[i].push_back(x);
      mark[i][x] = true;
    } else {
      break;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= 5000; i++) {
    G[i].push_back(0);
    mark[i][0] = true;
  }
  ans.push_back(0);
  has[0] = true;
  for (int i = 1; i <= n; i++) {
    int ptr = 0;
    while (ptr < G[a[i]].size()) {
      int x = G[a[i]][ptr++] ^ a[i];
      if (!has[x]) {
        ans.push_back(x);
        has[x] = true;
      }
      add(a[i], x);
    }
    G[a[i]].clear();
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << "\n";
  for (auto x : ans) {
    cout << x << " ";
  }
  return 0;
}

```