#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
inline int qpow(int a, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a;
    }
    a *= a;
    n >>= 1;
  }
  return ans;
}
inline long long _qpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
inline long long ksm(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
const int maxnum = 1e6 + 5;
const int MAXN = (1 << 13) - 1;
set<int> st[maxnum];
int n;
int x;
int f[MAXN + 5];
void slove() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    st[x].insert(i);
  }
  for (int i = 0; i <= MAXN; i++) f[i] = 0x3f3f3f3f;
  f[0] = 0;
  for (int i = 0; i <= MAXN; i++) {
    for (int j = 0; j <= MAXN; j++) {
      auto it = st[i].upper_bound(f[j]);
      if (it != st[i].end()) {
        f[i ^ j] = min(f[i ^ j], *it);
      }
    }
  }
  vector<int> ans;
  for (int i = 0; i <= MAXN; i++) {
    if (f[i] != 0x3f3f3f3f) ans.push_back(i);
  }
  cout << ans.size() << "\n";
  for (int x : ans) cout << x << " ";
  cout << "\n";
}
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  ;
  slove();
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize(3)
using namespace std;
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
inline int qpow(int a, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) {
      ans *= a;
    }
    a *= a;
    n >>= 1;
  }
  return ans;
}
inline long long _qpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
inline long long ksm(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
const int maxnum = 1e6 + 5;
const int MAXN = (1 << 13) - 1;
set<int> st[maxnum];
int n;
int x;
int f[MAXN + 5];
void slove() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    st[x].insert(i);
  }
  for (int i = 0; i <= MAXN; i++) f[i] = 0x3f3f3f3f;
  f[0] = 0;
  for (int i = 0; i <= MAXN; i++) {
    for (int j = 0; j <= MAXN; j++) {
      auto it = st[i].upper_bound(f[j]);
      if (it != st[i].end()) {
        f[i ^ j] = min(f[i ^ j], *it);
      }
    }
  }
  vector<int> ans;
  for (int i = 0; i <= MAXN; i++) {
    if (f[i] != 0x3f3f3f3f) ans.push_back(i);
  }
  cout << ans.size() << "\n";
  for (int x : ans) cout << x << " ";
  cout << "\n";
}
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  ;
  slove();
  return 0;
}

```