#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T sig = 1;
  for (; !isdigit(c); c = getchar())
    if (c == '-') sig = -1;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= sig;
}
int dp[5001][8192]{};
class Solution {
 public:
  void solve() {
    int n;
    read(n);
    vector<int> a(n);
    vector<vector<int>> nums(5001);
    for (int i = 0; i < n; ++i) {
      read(a[i]);
      nums[a[i]].push_back(i + 1);
    }
    memset(dp, 0x3f, sizeof(dp));
    dp[0][0] = 0;
    vector<bool> can(8192);
    can[0] = true;
    for (int i = 1; i <= 5000; ++i) {
      for (int j = 1; j < 8192; ++j) {
        dp[i][j] = dp[i - 1][j];
        if (i == j && !nums[i].empty()) dp[i][j] = min(dp[i][j], nums[i][0]);
        auto it = upper_bound(nums[i].begin(), nums[i].end(), dp[i - 1][i ^ j]);
        if (it != nums[i].end()) dp[i][j] = min(dp[i][j], *it);
        if (dp[i][j] < 0x3f3f3f3f) can[j] = true;
      }
    }
    int ans = 0;
    for (int i = 0; i < 8192; ++i)
      if (can[i]) ans++;
    printf("%d\n", ans);
    for (int i = 0; i < 8192; ++i)
      if (can[i]) printf("%d ", i);
    printf("\n");
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution solution = Solution();
  solution.solve();
}

### Prompt

Create a solution in Cpp for 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 read(T &x) {
  x = 0;
  char c = getchar();
  T sig = 1;
  for (; !isdigit(c); c = getchar())
    if (c == '-') sig = -1;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= sig;
}
int dp[5001][8192]{};
class Solution {
 public:
  void solve() {
    int n;
    read(n);
    vector<int> a(n);
    vector<vector<int>> nums(5001);
    for (int i = 0; i < n; ++i) {
      read(a[i]);
      nums[a[i]].push_back(i + 1);
    }
    memset(dp, 0x3f, sizeof(dp));
    dp[0][0] = 0;
    vector<bool> can(8192);
    can[0] = true;
    for (int i = 1; i <= 5000; ++i) {
      for (int j = 1; j < 8192; ++j) {
        dp[i][j] = dp[i - 1][j];
        if (i == j && !nums[i].empty()) dp[i][j] = min(dp[i][j], nums[i][0]);
        auto it = upper_bound(nums[i].begin(), nums[i].end(), dp[i - 1][i ^ j]);
        if (it != nums[i].end()) dp[i][j] = min(dp[i][j], *it);
        if (dp[i][j] < 0x3f3f3f3f) can[j] = true;
      }
    }
    int ans = 0;
    for (int i = 0; i < 8192; ++i)
      if (can[i]) ans++;
    printf("%d\n", ans);
    for (int i = 0; i < 8192; ++i)
      if (can[i]) printf("%d ", i);
    printf("\n");
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution solution = Solution();
  solution.solve();
}

```