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;
int main() {
  int n, i, j, k, sum, sum1, m, mi, ii, t;
  while (cin >> n) {
    int ab[234567];
    for (i = 0; i < n; i++) cin >> ab[i];
    sort(ab, ab + n);
    for (i = 0, sum = 0, k = 0; i < n; i++) {
      mi = ab[i] + 5;
      if (k == 0) {
        k = 1;
        for (ii = i + 1, sum1 = 1; ii < n; ii++) {
          if (mi >= ab[ii])
            sum1++;
          else {
            t = ii;
            break;
          }
        }
        if (sum1 > sum) sum = sum1;
      } else {
        if (mi >= ab[t]) {
          for (ii = t, sum1 = t - i; ii < n; ii++) {
            if (mi >= ab[ii]) {
              sum1++;
            } else {
              t = ii;
              break;
            }
          }
          if (sum1 > sum) sum = sum1;
        }
      }
      if (i + sum1 == n) break;
    }
    cout << sum << endl;
  }
  return 0;
}