#include <bits/stdc++.h>
using namespace std;
const int N = 8192;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
vector<int> v[N];
int n, m, a[maxn], b[N], st[N], rec[N], ans;
int main() {
  n = read();
  ans = st[0] = 1;
  for (int i = 0; i < N; i++) v[i].push_back(0), rec[i] = N - 1;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    for (auto x : v[a[i]]) {
      int u = x ^ a[i];
      if (!st[u]) st[u] = 1, ++ans;
      if (a[i] <= rec[u]) {
        for (int j = a[i] + 1; j <= rec[u]; j++) v[j].push_back(u);
        rec[u] = a[i] - 1;
      }
    }
    v[a[i]].clear();
  }
  printf("%d\n", ans);
  for (int i = 0; i < N; i++)
    if (st[i]) printf("%d ", i);
  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 = 8192;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
vector<int> v[N];
int n, m, a[maxn], b[N], st[N], rec[N], ans;
int main() {
  n = read();
  ans = st[0] = 1;
  for (int i = 0; i < N; i++) v[i].push_back(0), rec[i] = N - 1;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    for (auto x : v[a[i]]) {
      int u = x ^ a[i];
      if (!st[u]) st[u] = 1, ++ans;
      if (a[i] <= rec[u]) {
        for (int j = a[i] + 1; j <= rec[u]; j++) v[j].push_back(u);
        rec[u] = a[i] - 1;
      }
    }
    v[a[i]].clear();
  }
  printf("%d\n", ans);
  for (int i = 0; i < N; i++)
    if (st[i]) printf("%d ", i);
  return 0;
}

```