#include <bits/stdc++.h>
using namespace std;
const int N = 1000000 + 9;
int a[N], nxt[N], pos[N], vis[N], minv[N];
vector<int> vc[N];
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = n; i >= 1; --i) {
    nxt[i] = pos[a[i]];
    pos[a[i]] = i;
  }
  minv[0] = 0;
  for (int i = 1; i < (1 << 13); ++i) {
    minv[i] = 5000;
  }
  vis[0] = true;
  for (int i = 1; i <= 5000; ++i) {
    if (pos[i]) vc[pos[i]].push_back(0);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j : vc[i]) {
      int k = j ^ a[i];
      vis[k] = true;
      if (a[i] < minv[k]) {
        for (int x = a[i] + 1; x <= minv[k]; ++x) {
          if (pos[x]) vc[pos[x]].push_back(k);
        }
        minv[k] = a[i];
      }
    }
    pos[a[i]] = nxt[i];
  }
  vector<int> ans;
  for (int i = 0; i < (1 << 13); ++i) {
    if (vis[i]) ans.push_back(i);
  }
  printf("%d\n", (int)ans.size());
  for (int i : ans) printf("%d ", i);
  printf("\n");
}
int main() {
  int T = 1;
  while (T--) solve();
  return 0;
}

### Prompt

Generate 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 N = 1000000 + 9;
int a[N], nxt[N], pos[N], vis[N], minv[N];
vector<int> vc[N];
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = n; i >= 1; --i) {
    nxt[i] = pos[a[i]];
    pos[a[i]] = i;
  }
  minv[0] = 0;
  for (int i = 1; i < (1 << 13); ++i) {
    minv[i] = 5000;
  }
  vis[0] = true;
  for (int i = 1; i <= 5000; ++i) {
    if (pos[i]) vc[pos[i]].push_back(0);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j : vc[i]) {
      int k = j ^ a[i];
      vis[k] = true;
      if (a[i] < minv[k]) {
        for (int x = a[i] + 1; x <= minv[k]; ++x) {
          if (pos[x]) vc[pos[x]].push_back(k);
        }
        minv[k] = a[i];
      }
    }
    pos[a[i]] = nxt[i];
  }
  vector<int> ans;
  for (int i = 0; i < (1 << 13); ++i) {
    if (vis[i]) ans.push_back(i);
  }
  printf("%d\n", (int)ans.size());
  for (int i : ans) printf("%d ", i);
  printf("\n");
}
int main() {
  int T = 1;
  while (T--) solve();
  return 0;
}

```