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;
int main() {
  long long n, k;
  cin >> n;
  vector<long long> f, f1;
  long long a[n];
  for (long long i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  long long cnt = 1;
  for (long long i = 1; i < n; ++i) {
    if (a[i] == a[i - 1])
      cnt++;
    else {
      f.push_back(cnt);
      f1.push_back(a[i - 1]);
      cnt = 1;
    }
  }
  f.push_back(cnt);
  f1.push_back(a[n - 1]);
  long long c = f[0], mx = -1, s = a[0], i = 1, j = 0;
  long long p = 0;
  while (i < f1.size() && j < f1.size()) {
    if (f1[i] - s <= 5)
      c += f[i], i++;
    else
      mx = max(mx, c), j++, c = f[j], s = f1[j], i = j + 1;
    p++;
  }
  mx = max(mx, c);
  cout << mx;
  return 0;
}