#include <bits/stdc++.h>
using namespace std;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 1000000;
const int NBIT = 13;
const int MAXA = 5000;
int n;
int a[MAXN];
vector<int> ans;
int best[1 << NBIT];
vector<int> opt[MAXA + 1];
void solve() {
  for (int i = (0); i < (1 << NBIT); ++i) best[i] = INT_MAX;
  auto relax = [&](int at, int mx) {
    if (mx >= best[at]) return;
    for (int i = (mx + 1); i <= (min(best[at], MAXA)); ++i)
      opt[i].push_back(at);
    best[at] = mx;
  };
  relax(0, -1);
  for (int i = (0); i < (n); ++i) {
    int cur = a[i];
    while (((int)(opt[cur]).size()) > 0) {
      int at = opt[cur].back();
      opt[cur].pop_back();
      assert(best[at] < cur);
      relax(at ^ cur, cur);
    }
  }
  ans.clear();
  for (int i = (0); i < (1 << NBIT); ++i)
    if (best[i] != INT_MAX) ans.push_back(i);
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  solve();
  printf("%d\n", ((int)(ans).size()));
  for (int i = (0); i < (((int)(ans).size())); ++i) {
    if (i != 0) printf(" ");
    printf("%d", ans[i]);
  }
  puts("");
}
int main() {
  run();
  return 0;
}

### 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;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 1000000;
const int NBIT = 13;
const int MAXA = 5000;
int n;
int a[MAXN];
vector<int> ans;
int best[1 << NBIT];
vector<int> opt[MAXA + 1];
void solve() {
  for (int i = (0); i < (1 << NBIT); ++i) best[i] = INT_MAX;
  auto relax = [&](int at, int mx) {
    if (mx >= best[at]) return;
    for (int i = (mx + 1); i <= (min(best[at], MAXA)); ++i)
      opt[i].push_back(at);
    best[at] = mx;
  };
  relax(0, -1);
  for (int i = (0); i < (n); ++i) {
    int cur = a[i];
    while (((int)(opt[cur]).size()) > 0) {
      int at = opt[cur].back();
      opt[cur].pop_back();
      assert(best[at] < cur);
      relax(at ^ cur, cur);
    }
  }
  ans.clear();
  for (int i = (0); i < (1 << NBIT); ++i)
    if (best[i] != INT_MAX) ans.push_back(i);
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  solve();
  printf("%d\n", ((int)(ans).size()));
  for (int i = (0); i < (((int)(ans).size())); ++i) {
    if (i != 0) printf(" ");
    printf("%d", ans[i]);
  }
  puts("");
}
int main() {
  run();
  return 0;
}

```