Problem: You are a coach at your local university. There are n students under your supervision, the programming skill of the i-th student is a_i.

You have to create a team for a new programming competition. As you know, the more students some team has the more probable its victory is! So you have to create a team with the maximum number of students. But you also know that a team should be balanced. It means that the programming skill of each pair of students in a created team should differ by no more than 5.

Your task is to report the maximum possible number of students in a balanced team.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of students.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is a programming skill of the i-th student.

Output

Print one integer — the maximum possible number of students in a balanced team.

Examples

Input


6
1 10 17 12 15 2


Output


3


Input


10
1337 1337 1337 1337 1337 1337 1337 1337 1337 1337


Output


10


Input


6
1 1000 10000 10 100 1000000000


Output


1

Note

In the first example you can create a team with skills [12, 17, 15].

In the second example you can take all students in a team because their programming skills are equal.

In the third example you can create a team consisting of a single student (and you cannot create a team consisting of at least two students).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
long long int power(long long int a, long long int b, long long int m) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<long long int> v;
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    v.push_back(x);
  }
  sort((v).begin(), (v).end());
  long long int ans = 1;
  for (long long int i = 0; i < n; i++) {
    long long int x = v[i];
    x += 5;
    long long int ind = upper_bound((v).begin(), (v).end(), x) - v.begin();
    ind--;
    ans = max(ans, ind - i + 1);
  }
  cout << ans << endl;
}