#include <bits/stdc++.h>
using namespace std;
const int kMax = 1 << 13;
vector<int> g[kMax];
int r[kMax];
bitset<kMax> vis;
void TestCase() {
  int n;
  cin >> n;
  for (int x = 1; x < kMax; ++x) {
    g[x].emplace_back(0);
    r[x] = kMax;
  }
  vis[0] = true;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    for (int sxor : g[x]) {
      int new_xor = sxor ^ x;
      vis[new_xor] = true;
      while (x < r[new_xor]) {
        --r[new_xor];
        if (x < r[new_xor]) {
          g[r[new_xor]].emplace_back(new_xor);
        }
      }
    }
    g[x].clear();
  }
  cout << vis.count() << '\n';
  for (int x = 0; x < kMax; ++x) {
    if (vis[x]) {
      cout << x << ' ';
    }
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests = 1;
  for (int tc = 1; tc <= tests; ++tc) {
    TestCase();
  }
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
const int kMax = 1 << 13;
vector<int> g[kMax];
int r[kMax];
bitset<kMax> vis;
void TestCase() {
  int n;
  cin >> n;
  for (int x = 1; x < kMax; ++x) {
    g[x].emplace_back(0);
    r[x] = kMax;
  }
  vis[0] = true;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    for (int sxor : g[x]) {
      int new_xor = sxor ^ x;
      vis[new_xor] = true;
      while (x < r[new_xor]) {
        --r[new_xor];
        if (x < r[new_xor]) {
          g[r[new_xor]].emplace_back(new_xor);
        }
      }
    }
    g[x].clear();
  }
  cout << vis.count() << '\n';
  for (int x = 0; x < kMax; ++x) {
    if (vis[x]) {
      cout << x << ' ';
    }
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests = 1;
  for (int tc = 1; tc <= tests; ++tc) {
    TestCase();
  }
  return 0;
}

```