#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void _do(T x) {
  cerr << x << "\n";
}
template <typename T, typename... U>
void _do(T x, U... y) {
  cerr << x << ", ";
  _do(y...);
}
const int MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const long long INF = 3e18;
const long double PI = 3.14159265358979323846;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fpow(long long a, long long b, long long m) {
  if (!b) return 1;
  long long ans = fpow(a * a % m, b / 2, m);
  return (b % 2 ? ans * a % m : ans);
}
long long inv(long long a, long long m) { return fpow(a, m - 2, m); }
const long long K = 8192;
long long a[1000005], cur[K];
vector<long long> ed[5005];
bool can[K];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 0; i < 5001; i++) ed[i].push_back(0);
  for (long long i = 0; i < K; i++) cur[i] = 5001;
  can[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long x : ed[a[i]]) {
      long long y = x ^ a[i];
      can[y] = 1;
      while (cur[y] > a[i]) {
        cur[y]--;
        if (cur[y] != a[i]) ed[cur[y]].push_back(y);
      }
    }
    ed[a[i]].clear();
  }
  vector<long long> ans;
  for (long long i = 0; i < K; i++)
    if (can[i]) ans.push_back(i);
  cout << ans.size() << "\n";
  for (long long u : ans) cout << u << " ";
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void _do(T x) {
  cerr << x << "\n";
}
template <typename T, typename... U>
void _do(T x, U... y) {
  cerr << x << ", ";
  _do(y...);
}
const int MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const long long INF = 3e18;
const long double PI = 3.14159265358979323846;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fpow(long long a, long long b, long long m) {
  if (!b) return 1;
  long long ans = fpow(a * a % m, b / 2, m);
  return (b % 2 ? ans * a % m : ans);
}
long long inv(long long a, long long m) { return fpow(a, m - 2, m); }
const long long K = 8192;
long long a[1000005], cur[K];
vector<long long> ed[5005];
bool can[K];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 0; i < 5001; i++) ed[i].push_back(0);
  for (long long i = 0; i < K; i++) cur[i] = 5001;
  can[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long x : ed[a[i]]) {
      long long y = x ^ a[i];
      can[y] = 1;
      while (cur[y] > a[i]) {
        cur[y]--;
        if (cur[y] != a[i]) ed[cur[y]].push_back(y);
      }
    }
    ed[a[i]].clear();
  }
  vector<long long> ans;
  for (long long i = 0; i < K; i++)
    if (can[i]) ans.push_back(i);
  cout << ans.size() << "\n";
  for (long long u : ans) cout << u << " ";
}

```