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 val = 0;
int binsearch(int l, int r, int arr[]) {
  if (l > r) {
    return -1;
  }
  int mid = (l + r) / 2;
  if (l == r) {
    if (arr[mid] > val) {
      return -1;
    } else {
      return mid;
    }
  }
  if (arr[mid] <= val && arr[mid + 1] <= val) {
    return binsearch(mid + 1, r, arr);
  } else if (arr[mid] <= val && arr[mid + 1] > val) {
    return mid;
  } else if (arr[mid] > val) {
    return binsearch(l, mid - 1, arr);
  }
}
int main() {
  int n;
  cin >> n;
  int i, arr[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  if (n == 1) {
    cout << 1;
    return 0;
  }
  int max = 0;
  for (i = 0; i < n; i++) {
    val = arr[i] + 5;
    if (val >= arr[n - 1]) {
      if (max < n - i) {
        max = n - i;
      }
      break;
    }
    int k = binsearch(i + 1, n - 1, arr);
    if (k == -1) {
      if (max < 1) {
        max = 1;
      }
    }
    if (max < k - i + 1) {
      max = k - i + 1;
    }
  }
  cout << max;
  return 0;
}