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 N = 5e3 + 10;
int a[N], dp[N][N], ind[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    int tmp = i;
    while (tmp >= 0 && a[i] - a[tmp] <= 5) {
      ind[i] = tmp;
      tmp--;
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 1; j <= min(i + 1, k); j++) {
      if (i) dp[i][j] = dp[i - 1][j];
      dp[i][j] =
          max(dp[i][j],
              i - ind[i] + 1 + (ind[i] - 1 >= 0 ? dp[ind[i] - 1][j - 1] : 0));
    }
  int ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = 1; j <= k; j++) ans = max(dp[i][j], ans);
  cout << ans << "\n";
  return 0;
}