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 v[200007];
int main() {
  int n;
  map<int, int> m;
  while (~scanf("%d", &n)) {
    int ans = 0;
    m.clear();
    int a;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a);
      if (m[a] == 0) {
        m[a] = 1;
        ans++;
        v[ans] = a;
      } else {
        m[a]++;
      }
    }
    std::sort(v + 1, v + ans + 1);
    int max1 = 0;
    for (int i = 1; i <= ans; i++) {
      int temp = m[v[i]] + m[v[i] + 1] + m[v[i] + 2] + m[v[i] + 3] +
                 m[v[i] + 4] + m[v[i] + 5];
      if (temp > max1) {
        max1 = temp;
      }
    }
    printf("%d\n", max1);
  }
}