#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
int n;
int a[1000010];
vector<int> vec[5010];
int dp[2][1000010];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    vec[a[i]].push_back(i);
  }
  int per = 0, now = 1;
  for (int i = 1; i < 9000; ++i) dp[per][i] = 0x3f3f3f3f;
  dp[per][0] = 0;
  for (int i = 1; i <= 5000; ++i) {
    for (int j = 0; j < 9000; ++j) {
      dp[now][j] = dp[per][j];
    }
    for (int j = 0; j < 9000; ++j) {
      if (dp[per][j] == 0x3f3f3f3f) continue;
      auto it = upper_bound(vec[i].begin(), vec[i].end(), dp[per][j]);
      if (it != vec[i].end()) {
        dp[now][j ^ i] = min(dp[now][j ^ i], *it);
      }
    }
    now++;
    per++;
    now %= 2;
    per %= 2;
  }
  int cnt = 0;
  for (int j = 0; j < 9000; ++j) {
    if (dp[per][j] != 0x3f3f3f3f) {
      cnt++;
    }
  }
  printf("%d\n", cnt);
  for (int j = 0; j < 9000; ++j) {
    if (dp[per][j] != 0x3f3f3f3f) {
      printf("%d ", j);
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 long long mod = 1e9 + 7;
int n;
int a[1000010];
vector<int> vec[5010];
int dp[2][1000010];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    vec[a[i]].push_back(i);
  }
  int per = 0, now = 1;
  for (int i = 1; i < 9000; ++i) dp[per][i] = 0x3f3f3f3f;
  dp[per][0] = 0;
  for (int i = 1; i <= 5000; ++i) {
    for (int j = 0; j < 9000; ++j) {
      dp[now][j] = dp[per][j];
    }
    for (int j = 0; j < 9000; ++j) {
      if (dp[per][j] == 0x3f3f3f3f) continue;
      auto it = upper_bound(vec[i].begin(), vec[i].end(), dp[per][j]);
      if (it != vec[i].end()) {
        dp[now][j ^ i] = min(dp[now][j ^ i], *it);
      }
    }
    now++;
    per++;
    now %= 2;
    per %= 2;
  }
  int cnt = 0;
  for (int j = 0; j < 9000; ++j) {
    if (dp[per][j] != 0x3f3f3f3f) {
      cnt++;
    }
  }
  printf("%d\n", cnt);
  for (int j = 0; j < 9000; ++j) {
    if (dp[per][j] != 0x3f3f3f3f) {
      printf("%d ", j);
    }
  }
  return 0;
}

```