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:
import sys
from collections import Counter
def i_ints():
    return map(int, sys.stdin.readline().split())

n, k = i_ints()
c = Counter(i_ints())
c2 = dict()

a = sorted(c)
for i in sorted(a):
    c2[i] = sum(c[j] for j in range(i, i + 6))

# a are all possible levels of students
# c2[i] is maximal group size where lowest level is i
    
len_a = len(a)
next_group = [-1] * len(a)
for i in range(len_a):
    for j in range(i + 1, len_a):
        if a[j] > a[i] + 5:
            next_group[i] = j
            break
        
# if a group starts with i-th element, 
# then the next possible group starts with next_group[i]-th element

maxes = [0] * n  # for a maximum of 0 groups
for ii in range(k):
    old_maxes = maxes  
    old_maxes.append(0)  # access where next_group[...] == -1
    maxes = []
    # max number of groups, try to find better maxes each round
    for i, aa in enumerate(a):
        maxes.append(c2[a[i]] + old_maxes[next_group[i]])
    m = 0
    for i in range(len(a)-1, -1, -1):
        if maxes[i] > m:
            m = maxes[i]
        else:
            maxes[i] = m
        
print(max(maxes))