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 PYTHON3):

Solution:
from sys import stdin
from collections import defaultdict, Counter
from bisect import bisect_left
from math import sqrt
from heapq import *
###############################################################
def iinput(): return int(stdin.readline())
def minput(): return map(int, stdin.readline().split())
def linput(): return list(map(int, stdin.readline().split()))
###############################################################


t = 1
while t:
    t -= 1
    n = iinput()
    a = linput()
    a.sort()
    i, j = 0, 1
    mx = 1
    diff = 0
    while i<n and j<n:
        diff = a[j] - a[i]
        if diff <= 5:
            mx = max(mx, j-i+1)
            j += 1
        else: i += 1
    print(mx)