#include <bits/stdc++.h>
using namespace std;
const long double PI = acos(-1.0);
const long double EPS = 1e-7;
const int MOD = 1e9 + 7;
int POW(int b, int e) {
  int r = 1;
  while (e) {
    if (e % 2 == 1) {
      r *= b;
      r %= MOD;
    }
    e /= 2;
    b *= b;
    b %= MOD;
  }
  return r;
}
int gcd(int a, int b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
int INV(int a) { return POW(a, MOD - 2); }
vector<vector<bool>> table(5001, vector<bool>(8192, false));
vector<vector<int>> upd(5001);
void setv(int a, int b) {
  if (!table[a][b]) {
    table[a][b] = true;
    if (a) {
      setv(a - 1, b);
    }
    upd[a].push_back(b);
  }
}
void roll(int a) {
  while (!upd[a].empty()) {
    setv(a - 1, upd[a].back() ^ a);
    upd[a].pop_back();
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i <= 5000; i++) {
    setv(i, 0);
  }
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] > 0) roll(a[i]);
  }
  vector<int> out;
  for (int i = 0; i < 8192; i++) {
    if (table[0][i]) {
      out.push_back(i);
    }
  }
  cout << (int)out.size() << '\n';
  for (auto v : out) {
    cout << v << ' ';
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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 long double PI = acos(-1.0);
const long double EPS = 1e-7;
const int MOD = 1e9 + 7;
int POW(int b, int e) {
  int r = 1;
  while (e) {
    if (e % 2 == 1) {
      r *= b;
      r %= MOD;
    }
    e /= 2;
    b *= b;
    b %= MOD;
  }
  return r;
}
int gcd(int a, int b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
int INV(int a) { return POW(a, MOD - 2); }
vector<vector<bool>> table(5001, vector<bool>(8192, false));
vector<vector<int>> upd(5001);
void setv(int a, int b) {
  if (!table[a][b]) {
    table[a][b] = true;
    if (a) {
      setv(a - 1, b);
    }
    upd[a].push_back(b);
  }
}
void roll(int a) {
  while (!upd[a].empty()) {
    setv(a - 1, upd[a].back() ^ a);
    upd[a].pop_back();
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i <= 5000; i++) {
    setv(i, 0);
  }
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] > 0) roll(a[i]);
  }
  vector<int> out;
  for (int i = 0; i < 8192; i++) {
    if (table[0][i]) {
      out.push_back(i);
    }
  }
  cout << (int)out.size() << '\n';
  for (auto v : out) {
    cout << v << ' ';
  }
}

```