#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  long long n;
  cin >> n;
  vector<long long> a(n);
  map<long long, long long> mp;
  vector<long long> cnt(n);
  for (long long i = 0; i < n; ++i) {
    long long o;
    cin >> o;
    a[i] = o;
    mp[o]++;
    cnt[i] = mp[o];
  }
  vector<long long> n_a;
  for (long long i1 = 0; i1 < n; ++i1) {
    if (cnt[i1] == 1 || cnt[i1] == mp[a[i1]] || cnt[i1] == 2 ||
        cnt[i1] == mp[a[i1]] - 1) {
      n_a.push_back(a[i1]);
    }
  }
  a = n_a;
  n = a.size();
  long long mx = 8191;
  bool dp[n + 1][mx + 1];
  for (long long j = 0; j < n + 1; ++j) {
    for (long long i = 0; i < mx + 1; ++i) {
      dp[j][i] = false;
    }
  }
  dp[0][0] = true;
  long long last[mx + 1];
  for (long long m = 0; m < mx + 1; ++m) {
    last[m] = inf;
  }
  last[0] = -inf;
  for (long long k = 1; k < n + 1; ++k) {
    for (long long i = 0; i < mx + 1; ++i) {
      long long need = i ^ a[k - 1];
      if (last[need] < a[k - 1]) {
        dp[k][i] = true;
      }
    }
    for (long long j = 0; j < mx + 1; ++j) {
      if (dp[k][j]) last[j] = min(last[j], a[k - 1]);
    }
  }
  vector<bool> used(mx + 1);
  for (long long k1 = 0; k1 <= n; ++k1) {
    for (long long l = 0; l < mx + 1; ++l) {
      if (dp[k1][l]) used[l] = true;
    }
  }
  long long cc = 0;
  for (long long l1 = 0; l1 < mx + 1; ++l1) {
    cc += used[l1];
  }
  cout << cc << "\n";
  for (long long l1 = 0; l1 < mx + 1; ++l1) {
    if (used[l1]) {
      cout << l1 << " ";
    }
  }
  cout << "\n";
}

### Prompt

Please create a solution in Cpp 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 long long inf = 1e18;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  long long n;
  cin >> n;
  vector<long long> a(n);
  map<long long, long long> mp;
  vector<long long> cnt(n);
  for (long long i = 0; i < n; ++i) {
    long long o;
    cin >> o;
    a[i] = o;
    mp[o]++;
    cnt[i] = mp[o];
  }
  vector<long long> n_a;
  for (long long i1 = 0; i1 < n; ++i1) {
    if (cnt[i1] == 1 || cnt[i1] == mp[a[i1]] || cnt[i1] == 2 ||
        cnt[i1] == mp[a[i1]] - 1) {
      n_a.push_back(a[i1]);
    }
  }
  a = n_a;
  n = a.size();
  long long mx = 8191;
  bool dp[n + 1][mx + 1];
  for (long long j = 0; j < n + 1; ++j) {
    for (long long i = 0; i < mx + 1; ++i) {
      dp[j][i] = false;
    }
  }
  dp[0][0] = true;
  long long last[mx + 1];
  for (long long m = 0; m < mx + 1; ++m) {
    last[m] = inf;
  }
  last[0] = -inf;
  for (long long k = 1; k < n + 1; ++k) {
    for (long long i = 0; i < mx + 1; ++i) {
      long long need = i ^ a[k - 1];
      if (last[need] < a[k - 1]) {
        dp[k][i] = true;
      }
    }
    for (long long j = 0; j < mx + 1; ++j) {
      if (dp[k][j]) last[j] = min(last[j], a[k - 1]);
    }
  }
  vector<bool> used(mx + 1);
  for (long long k1 = 0; k1 <= n; ++k1) {
    for (long long l = 0; l < mx + 1; ++l) {
      if (dp[k1][l]) used[l] = true;
    }
  }
  long long cc = 0;
  for (long long l1 = 0; l1 < mx + 1; ++l1) {
    cc += used[l1];
  }
  cout << cc << "\n";
  for (long long l1 = 0; l1 < mx + 1; ++l1) {
    if (used[l1]) {
      cout << l1 << " ";
    }
  }
  cout << "\n";
}

```