#include <bits/stdc++.h>
using namespace std;
const int MAXN = (1 << 13) - 1;
int r[MAXN + 1], ans[MAXN + 1];
vector<int> g[MAXN + 1];
inline void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i <= MAXN; i++) {
    g[i].push_back(0);
    r[i] = MAXN;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    for (int j = 0; j < g[x].size(); j++) {
      int to = g[x][j] ^ x;
      ans[to] = 1;
      while (r[to] > x) {
        g[r[to]].push_back(to);
        r[to]--;
      }
    }
    g[x].clear();
  }
  int cnt = 0;
  ans[0] = 1;
  for (int i = 0; i <= MAXN; i++) {
    if (ans[i] == 1) cnt++;
  }
  printf("%d\n", cnt);
  for (int i = 0; i <= MAXN; i++) {
    if (i == 0)
      printf("%d", i);
    else if (ans[i] == 1)
      printf(" %d", i);
  }
  printf("\n");
}
int main() { solve(); }

### 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;
const int MAXN = (1 << 13) - 1;
int r[MAXN + 1], ans[MAXN + 1];
vector<int> g[MAXN + 1];
inline void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i <= MAXN; i++) {
    g[i].push_back(0);
    r[i] = MAXN;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    for (int j = 0; j < g[x].size(); j++) {
      int to = g[x][j] ^ x;
      ans[to] = 1;
      while (r[to] > x) {
        g[r[to]].push_back(to);
        r[to]--;
      }
    }
    g[x].clear();
  }
  int cnt = 0;
  ans[0] = 1;
  for (int i = 0; i <= MAXN; i++) {
    if (ans[i] == 1) cnt++;
  }
  printf("%d\n", cnt);
  for (int i = 0; i <= MAXN; i++) {
    if (i == 0)
      printf("%d", i);
    else if (ans[i] == 1)
      printf(" %d", i);
  }
  printf("\n");
}
int main() { solve(); }

```