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 A[5005], n, k, L[5005][5005], S[5005], E[5005];
void doc() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
  }
}
void xuly() {
  sort(A + 1, A + n + 1);
  for (int i = 1; i <= n; i++) {
    S[i] = 1;
    E[i] = i;
    for (int j = i - 1; j > 0; j--)
      if (A[i] - A[j] <= 5) {
        S[i]++;
        E[i] = j;
      } else
        break;
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      L[i][j] = max(max(L[i][j - 1], L[i][j]), L[i - 1][E[j] - 1] + S[j]);
    }
  }
  cout << L[k][n];
}
int main() {
  doc();
  xuly();
  return 0;
}