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>
int matr[200010][6];
int mas[200010];
int counts[200010];
int n;
void run(int a, int d) {
  int checkRes = 0;
  int buf;
  matr[a][d] = 1;
  int i = a + 1;
  while (mas[i] - mas[a] <= d && i < n) {
    buf = d - (mas[i] - mas[a]);
    if (matr[i][buf] == 0) {
      run(i, buf);
    }
    if (checkRes < matr[i][buf]) {
      checkRes = matr[i][buf];
    }
    ++i;
  }
  matr[a][d] += checkRes;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &mas[i]);
  }
  std::sort(&mas[0], &mas[n]);
  int l = 1;
  int s = 0;
  for (int i = 1; i < n; ++i) {
    if (mas[i] - mas[s] < 6) {
      ++l;
    } else {
      do {
        counts[s] = l;
        ++s;
        --l;
      } while (mas[i] - mas[s] > 5);
      ++l;
    }
  }
  counts[s] = l;
  int res = counts[0];
  for (int i = 1; i < n; ++i) {
    if (counts[i] > res) res = counts[i];
  }
  printf("%d", res);
  return 0;
}