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;
const int MAXN = 200005;
int n, k;
int a[MAXN];
deque<pair<int, int> > maxn, minn;
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + 1 + n);
  int tmp = 0, ans = 0;
  for (int i = 1; i <= n; ++i) {
    while (minn.size() && minn.back().first > a[i]) minn.pop_back();
    minn.push_back(make_pair(a[i], i));
    while (maxn.size() && maxn.back().first < a[i]) maxn.pop_back();
    maxn.push_back(make_pair(a[i], i));
    while (maxn.front().first - minn.front().first > 5) {
      tmp = min(maxn.front().second, minn.front().second);
      if (maxn.front().second < minn.front().second)
        maxn.pop_front();
      else
        minn.pop_front();
    }
    ans = max(ans, i - tmp);
  }
  printf("%d\n", ans);
  return 0;
}