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;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  int ptr1 = 0, ptr2 = 0;
  vector<int> ans;
  bool ok = false;
  for (int i = 0; i < n - 1; i++) {
    if (arr[i + 1] - arr[i] <= 5) {
      ok = true;
      break;
    }
  }
  if (!ok) {
    cout << 1;
    return 0;
  }
  bool flag = false;
  while (ptr2 != n - 1 && !flag) {
    if (abs(arr[ptr1] - arr[ptr2 + 1]) <= 5) {
      ptr2++;
      ans.push_back(abs(ptr2 - ptr1) + 1);
    } else {
      while (abs(arr[ptr1] - arr[ptr2 + 1]) > 5 &&
             abs(arr[ptr1] - arr[ptr2]) >= 0) {
        ptr1++;
        ans.push_back(abs(ptr2 - ptr1) + 1);
      }
      ans.push_back(abs(ptr2 - ptr1) + 1);
    }
  }
  sort(ans.begin(), ans.end());
  int final = ans[ans.size() - 1];
  cout << final;
  return 0;
}