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 MAX_N = 5e3 + 5;
int a[MAX_N];
int last[MAX_N];
int f[MAX_N][MAX_N];
int main() {
  int n, k;
  while (cin >> n >> k) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
    }
    sort(a + 1, a + 1 + n);
    for (int i = 1; i <= n; i++) {
      if (a[i] == a[i - 1]) {
        last[i] = last[i - 1];
      } else {
        for (int j = i; j <= n && a[j] <= a[i] + 5; j++) {
          last[i] = j;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= k; j++) {
        f[i + 1][j] = max(f[i + 1][j], f[i][j]);
        f[last[i + 1]][j + 1] =
            max(f[last[i + 1]][j + 1], f[i][j] + last[i + 1] - i);
      }
    }
    cout << f[n][k] << endl;
  }
  return 0;
}