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 pre[5005], dp[5005][5005];
int a[5005];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + n + 1);
  deque<int> dq;
  dq.push_front(1);
  for (int i = 2; i <= n; i++) {
    while (dq.size() && a[dq.front()] - a[dq.back()] > 5) {
      dq.pop_back();
    }
    pre[dq.front()] = dq.size();
    dq.push_front(i);
  }
  while (dq.size() && a[dq.front()] - a[dq.back()] > 5) {
    dq.pop_back();
  }
  pre[n] = dq.size();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      dp[i][j] = max(dp[i - 1][j], dp[i - pre[i]][j - 1] + pre[i]);
    }
  }
  printf("%d", dp[n][k]);
}