#include <bits/stdc++.h>
const int N = 1000005;
const int M = 10000;
int n, t;
inline void read(int &x) {
  register char ch = 0;
  register bool w = 0;
  for (x = 0; !std::isdigit(ch); w |= ch == '-', ch = getchar())
    ;
  for (; std::isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  (w) && (x = -x);
  return;
}
std::vector<int> reach[M];
int min0[M], vis[M], res[M];
void solve() {
  read(n);
  static int ai[N];
  for (int i = 1; i <= n; ++i) read(ai[i]);
  for (int i = 0; i < 8192; ++i) reach[i].push_back(0), min0[i] = 8192;
  for (int i = 1; i <= n; ++i) {
    if (vis[ai[i]]) continue;
    vis[ai[i]] = 1;
    for (auto v : reach[ai[i]]) {
      int xr = v ^ ai[i];
      for (int j = ai[i] + 1; j < min0[xr]; ++j) {
        if (vis[j]) std::vector<int>().swap(reach[j]), vis[j] = 0;
        reach[j].push_back(xr);
      }
      res[xr] = 1;
      min0[xr] = std::min(min0[xr], ai[i] + 1);
    }
  }
  res[0] = 1;
  std::vector<int> vec;
  for (int i = 0; i < 8192; ++i)
    if (res[i]) vec.push_back(i);
  printf("%lu\n", vec.size());
  for (auto v : vec) printf("%d ", v);
  printf("\n");
}
int main() {
  t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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>
const int N = 1000005;
const int M = 10000;
int n, t;
inline void read(int &x) {
  register char ch = 0;
  register bool w = 0;
  for (x = 0; !std::isdigit(ch); w |= ch == '-', ch = getchar())
    ;
  for (; std::isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  (w) && (x = -x);
  return;
}
std::vector<int> reach[M];
int min0[M], vis[M], res[M];
void solve() {
  read(n);
  static int ai[N];
  for (int i = 1; i <= n; ++i) read(ai[i]);
  for (int i = 0; i < 8192; ++i) reach[i].push_back(0), min0[i] = 8192;
  for (int i = 1; i <= n; ++i) {
    if (vis[ai[i]]) continue;
    vis[ai[i]] = 1;
    for (auto v : reach[ai[i]]) {
      int xr = v ^ ai[i];
      for (int j = ai[i] + 1; j < min0[xr]; ++j) {
        if (vis[j]) std::vector<int>().swap(reach[j]), vis[j] = 0;
        reach[j].push_back(xr);
      }
      res[xr] = 1;
      min0[xr] = std::min(min0[xr], ai[i] + 1);
    }
  }
  res[0] = 1;
  std::vector<int> vec;
  for (int i = 0; i < 8192; ++i)
    if (res[i]) vec.push_back(i);
  printf("%lu\n", vec.size());
  for (auto v : vec) printf("%d ", v);
  printf("\n");
}
int main() {
  t = 1;
  while (t--) solve();
  return 0;
}

```