#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100, M = 1 << 13;
int n, a[N], MIN[M], f[M];
vector<int> p[M];
inline void ckmin(int &a, int b) { a = ((a < b) ? a : b); }
inline void ckmax(int &a, int b) { a = ((a > b) ? a : b); }
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read(), p[a[i]].push_back(i);
  memset(MIN, 0x3f, sizeof(MIN));
  MIN[0] = 0;
  for (int i = 1; i <= 5000; i++)
    if (!p[i].empty()) {
      vector<int> b = p[i];
      memset(f, 0x3f, sizeof(f));
      for (int j = 0; j < M; j++)
        if (MIN[j] != 0x3f3f3f3f) {
          int pos = lower_bound(b.begin(), b.end(), MIN[j]) - b.begin();
          if (pos == (int)b.size())
            f[j ^ i] = 0x3f3f3f3f;
          else
            f[j ^ i] = b[pos];
        }
      for (int j = 0; j < M; j++) ckmin(MIN[j], f[j]);
    }
  vector<int> ans;
  for (int i = 0; i < M; i++)
    if (MIN[i] != 0x3f3f3f3f) ans.push_back(i);
  printf("%d\n", (int)ans.size());
  for (int i : ans) printf("%d ", i);
  printf("\n");
}

### 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 N = 1e6 + 100, M = 1 << 13;
int n, a[N], MIN[M], f[M];
vector<int> p[M];
inline void ckmin(int &a, int b) { a = ((a < b) ? a : b); }
inline void ckmax(int &a, int b) { a = ((a > b) ? a : b); }
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read(), p[a[i]].push_back(i);
  memset(MIN, 0x3f, sizeof(MIN));
  MIN[0] = 0;
  for (int i = 1; i <= 5000; i++)
    if (!p[i].empty()) {
      vector<int> b = p[i];
      memset(f, 0x3f, sizeof(f));
      for (int j = 0; j < M; j++)
        if (MIN[j] != 0x3f3f3f3f) {
          int pos = lower_bound(b.begin(), b.end(), MIN[j]) - b.begin();
          if (pos == (int)b.size())
            f[j ^ i] = 0x3f3f3f3f;
          else
            f[j ^ i] = b[pos];
        }
      for (int j = 0; j < M; j++) ckmin(MIN[j], f[j]);
    }
  vector<int> ans;
  for (int i = 0; i < M; i++)
    if (MIN[i] != 0x3f3f3f3f) ans.push_back(i);
  printf("%d\n", (int)ans.size());
  for (int i : ans) printf("%d ", i);
  printf("\n");
}

```