#include <bits/stdc++.h>
using namespace std;
inline int rd() {
  int x = 0;
  char ch, t = 0;
  while (!isdigit(ch = getchar())) t |= ch == '-';
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return x = t ? -x : x;
}
inline long long maxll(long long x, long long y) { return x > y ? x : y; }
inline long long minll(long long x, long long y) { return x < y ? x : y; }
inline long long absll(long long x) { return x > 0ll ? x : -x; }
inline long long gcd(long long x, long long y) {
  return (y == 0) ? x : gcd(y, x % y);
}
int n, All = 8191;
int a[1000005];
bitset<8200> ans[8200], vis[8200];
vector<int> pre[1000005];
int main() {
  n = rd();
  for (int i = 0; i <= All; i++) pre[i].push_back(0);
  for (int i = 1; i <= n; i++) {
    a[i] = rd();
    if (!vis[a[i]][a[i]]) pre[a[i]].push_back(a[i]), vis[a[i]][a[i]] = true;
  }
  for (int i = 1; i <= n; i++) {
    for (int j : pre[a[i]]) {
      ans[a[i]][j] = true;
      for (int k = a[i] + 1; k <= 5000; k++) {
        if (vis[k][k ^ j]) break;
        vis[k][k ^ j] = true, pre[k].push_back(k ^ j);
      }
    }
    pre[a[i]].clear();
  }
  for (int i = 1; i <= 5000; i++) ans[0] |= ans[i];
  printf("%lld\n", ans[0].count());
  for (int i = 0; i <= All; i++)
    if (ans[0][i]) printf("%d ", i);
  printf("\n");
  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;
inline int rd() {
  int x = 0;
  char ch, t = 0;
  while (!isdigit(ch = getchar())) t |= ch == '-';
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return x = t ? -x : x;
}
inline long long maxll(long long x, long long y) { return x > y ? x : y; }
inline long long minll(long long x, long long y) { return x < y ? x : y; }
inline long long absll(long long x) { return x > 0ll ? x : -x; }
inline long long gcd(long long x, long long y) {
  return (y == 0) ? x : gcd(y, x % y);
}
int n, All = 8191;
int a[1000005];
bitset<8200> ans[8200], vis[8200];
vector<int> pre[1000005];
int main() {
  n = rd();
  for (int i = 0; i <= All; i++) pre[i].push_back(0);
  for (int i = 1; i <= n; i++) {
    a[i] = rd();
    if (!vis[a[i]][a[i]]) pre[a[i]].push_back(a[i]), vis[a[i]][a[i]] = true;
  }
  for (int i = 1; i <= n; i++) {
    for (int j : pre[a[i]]) {
      ans[a[i]][j] = true;
      for (int k = a[i] + 1; k <= 5000; k++) {
        if (vis[k][k ^ j]) break;
        vis[k][k ^ j] = true, pre[k].push_back(k ^ j);
      }
    }
    pre[a[i]].clear();
  }
  for (int i = 1; i <= 5000; i++) ans[0] |= ans[i];
  printf("%lld\n", ans[0].count());
  for (int i = 0; i <= All; i++)
    if (ans[0][i]) printf("%d ", i);
  printf("\n");
  return 0;
}

```