#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chkMax(T &x, T y) {
  if (y > x) x = y;
}
template <typename T>
void chkMin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
const int N = 8200, INF = 0x3f3f3f3f;
int n, a[N], g[N];
vector<int> p[N];
int main() {
  read(n);
  for (int i = 1, x; i <= n; i++) read(x), p[x].push_back(i);
  memset(g, 0x3f, sizeof g);
  g[0] = 0;
  for (int x = 1; x <= 5000; x++) {
    if (!p[x].size()) continue;
    for (int i = 0; i < N; i++) {
      if (g[i] == INF) continue;
      if (g[i] < p[x].back()) {
        int pt = upper_bound(p[x].begin(), p[x].end(), g[i]) - p[x].begin();
        int ps = p[x][pt];
        if (ps < g[i ^ x]) {
          g[i ^ x] = ps;
        }
      }
    }
  }
  int c = 0;
  for (int i = 0; i < N; i++)
    if (g[i] != INF) c++;
  printf("%d\n", c);
  for (int i = 0; i < N; i++)
    if (g[i] != INF) printf("%d ", i);
  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;
template <typename T>
void chkMax(T &x, T y) {
  if (y > x) x = y;
}
template <typename T>
void chkMin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
const int N = 8200, INF = 0x3f3f3f3f;
int n, a[N], g[N];
vector<int> p[N];
int main() {
  read(n);
  for (int i = 1, x; i <= n; i++) read(x), p[x].push_back(i);
  memset(g, 0x3f, sizeof g);
  g[0] = 0;
  for (int x = 1; x <= 5000; x++) {
    if (!p[x].size()) continue;
    for (int i = 0; i < N; i++) {
      if (g[i] == INF) continue;
      if (g[i] < p[x].back()) {
        int pt = upper_bound(p[x].begin(), p[x].end(), g[i]) - p[x].begin();
        int ps = p[x][pt];
        if (ps < g[i ^ x]) {
          g[i ^ x] = ps;
        }
      }
    }
  }
  int c = 0;
  for (int i = 0; i < N; i++)
    if (g[i] != INF) c++;
  printf("%d\n", c);
  for (int i = 0; i < N; i++)
    if (g[i] != INF) printf("%d ", i);
  return 0;
}

```