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 PYTHON):

Solution:
#####################################   
import atexit, io, sys, collections, math, heapq, fractions
buffer = io.BytesIO() 
sys.stdout = buffer
@atexit.register 
def write(): sys.__stdout__.write(buffer.getvalue())   
####################################

n, k = map(int, raw_input().split())
ais = map(int, raw_input().split())
ais.sort()
dp = [0 for _ in range(n)]
prev = [0 for _ in range(n)]
for i in range(k):
	u = len(ais) - 1
	
	for j in range(len(ais)-1,-1,-1):
		dp[j] = 0
		if j +1 < len(ais):
			dp[j] = max(dp[j+1], dp[j])
		while(u >= 0 and ais[u] - ais[j] >5):
			u-=1
		dp[j] = max(dp[j], u - j +1+ (prev[u+1] if u+1 < len(ais) and i-1 >=0 else 0))
	prev = dp[::]

print max(dp)