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;
bool is_powertwo(long long n) {
  long long count = 0;
  while (n) {
    if ((n & 1) > 0) count++;
    n = n >> 1;
  }
  if (count == 1) return true;
  return false;
}
int main() {
  long long t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    for (long long i = 0; i < n; i++) cin >> arr[i];
    sort(arr, arr + n);
    long long maxm = 1;
    long long l = 0;
    long long r = 1;
    while (r < n) {
      if (arr[r] - arr[l] <= 5)
        r++;
      else {
        l++;
        r++;
      }
      maxm = max(maxm, r - l);
    }
    cout << maxm << endl;
  }
  return 0;
}