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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k, a;
  scanf("%d", &n);
  vector<int> nums;
  unordered_map<int, int> hashmap, sizes, parent, inp;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    nums.push_back(a);
  }
  sort(nums.begin(), nums.end());
  int smin = -1, smin_pos = 0, res = 1;
  for (int i = 0; i < n; i++) {
    a = nums[i];
    if (smin == -1) {
      smin = a;
      smin_pos = i;
    } else {
      if (a - smin <= 5) {
        continue;
      } else {
        res = max(res, i - smin_pos);
        while (a - nums[smin_pos] > 5 && smin_pos < i) smin_pos++;
        smin = nums[smin_pos];
      }
    }
  }
  if (nums[n - 1] - smin <= 5) {
    res = max(res, n - smin_pos);
  }
  printf("%d", res);
  return 0;
}