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:
"""
    This template is made by Satwik_Tiwari.
    python programmers can use this template :)) .
"""

#===============================================================================================
#importing some useful libraries.

import sys
import bisect
import heapq
from math import *
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#===============================================================================================
#some shortcuts

mod = pow(10, 9) + 7
def inp(): return sys.stdin.readline().strip() #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)


#===============================================================================================
# code here ;))

def solve():
    n = int(inp())
    a = sorted(lis())
    # print(a)
    ans = 0
    temp = 1
    f = False
    for i in range(0,n):
        if(f):
            break
        for j in range(temp,n):
            if(a[j] - a[i] < 6):
                if(j==n-1):
                    f = True
                    ans = max(j-i+1,ans)
                continue
            else:
                ans = max(j-i  , ans)
                temp = j
                break

    if(ans==0):
        print(1)
    else:
        print(ans)




testcase(1)