#include <bits/stdc++.h>
const int N = 1000100;
int n, k, a, dp[2][N], ans[N], cur;
std::vector<int> v[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 8192; ++i) dp[cur][i] = dp[cur ^ 1][i] = n + 1;
  dp[cur][0] = 0, k = 0;
  for (int i = 1; i <= n; ++i) scanf("%d", &a), v[a].push_back(i);
  for (int i = 0; i < 5000; ++i, cur ^= 1) {
    for (int j = 0; j < 8192; ++j) dp[cur ^ 1][j] = dp[cur][j];
    if (!v[i + 1].size()) continue;
    for (int j = 0; j < 8192; ++j) {
      if (dp[cur][j] == n + 1) continue;
      int id = std::lower_bound(v[i + 1].begin(), v[i + 1].end(), dp[cur][j]) -
               v[i + 1].begin();
      if (id == v[i + 1].size()) continue;
      dp[cur ^ 1][j ^ (i + 1)] =
          std::min(dp[cur ^ 1][j ^ (i + 1)], v[i + 1][id]);
    }
  }
  for (int i = 0; i < 8192; ++i)
    if (dp[cur][i] != n + 1) ans[++k] = i;
  printf("%d\n", k);
  for (int i = 1; i <= k; ++i) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

### Prompt

Please formulate 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>
const int N = 1000100;
int n, k, a, dp[2][N], ans[N], cur;
std::vector<int> v[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 8192; ++i) dp[cur][i] = dp[cur ^ 1][i] = n + 1;
  dp[cur][0] = 0, k = 0;
  for (int i = 1; i <= n; ++i) scanf("%d", &a), v[a].push_back(i);
  for (int i = 0; i < 5000; ++i, cur ^= 1) {
    for (int j = 0; j < 8192; ++j) dp[cur ^ 1][j] = dp[cur][j];
    if (!v[i + 1].size()) continue;
    for (int j = 0; j < 8192; ++j) {
      if (dp[cur][j] == n + 1) continue;
      int id = std::lower_bound(v[i + 1].begin(), v[i + 1].end(), dp[cur][j]) -
               v[i + 1].begin();
      if (id == v[i + 1].size()) continue;
      dp[cur ^ 1][j ^ (i + 1)] =
          std::min(dp[cur ^ 1][j ^ (i + 1)], v[i + 1][id]);
    }
  }
  for (int i = 0; i < 8192; ++i)
    if (dp[cur][i] != n + 1) ans[++k] = i;
  printf("%d\n", k);
  for (int i = 1; i <= k; ++i) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

```