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 d[5005][5005];
int a[5005], pre[5005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  a[0] = -1000000000;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j >= 0; j--) {
      if (a[j] + 5 < a[i]) {
        pre[i] = j;
        break;
      }
    }
  }
  int res = 0;
  for (int j = 1; j <= k; j++) {
    for (int i = 1; i <= n; i++) {
      if (j == 1) {
        d[i][j] = max(d[i - 1][j], i - pre[i]);
      } else {
        d[i][j] = max(d[i - 1][j], d[pre[i]][j - 1] + i - pre[i]);
      }
      res = max(res, d[i][j]);
    }
  }
  cout << res << endl;
  return 0;
}