#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
vector<int> g[10010];
int r[10010];
bool ans[10010];
const int M = (1 << 13) - 1;
int main() {
  int n = read();
  ans[0] = 1;
  for (register int i = 0; i <= M; ++i) g[i].push_back(0), r[i] = M;
  while (n--) {
    int x = read();
    for (int y : g[x]) {
      int to = x ^ y;
      ans[to] = 1;
      for (register int i = x + 1; i <= r[to]; ++i) g[i].push_back(to);
      r[to] = min(r[to], x);
    }
    g[x] = {};
  }
  int k = 0;
  for (register int i = 0; i <= M; ++i) k += ans[i];
  printf("%d\n", k);
  for (register int i = 0; i <= M; ++i)
    if (ans[i]) printf("%d ", i);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
vector<int> g[10010];
int r[10010];
bool ans[10010];
const int M = (1 << 13) - 1;
int main() {
  int n = read();
  ans[0] = 1;
  for (register int i = 0; i <= M; ++i) g[i].push_back(0), r[i] = M;
  while (n--) {
    int x = read();
    for (int y : g[x]) {
      int to = x ^ y;
      ans[to] = 1;
      for (register int i = x + 1; i <= r[to]; ++i) g[i].push_back(to);
      r[to] = min(r[to], x);
    }
    g[x] = {};
  }
  int k = 0;
  for (register int i = 0; i <= M; ++i) k += ans[i];
  printf("%d\n", k);
  for (register int i = 0; i <= M; ++i)
    if (ans[i]) printf("%d ", i);
  return 0;
}

```