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[5010];
int maxStart[5010];
int dp[5010][2];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a, a + n);
  int l = 0, r = 0;
  while (r < n) {
    if (a[r] - a[l] <= 5) {
      maxStart[r] = l;
      ++r;
      continue;
    }
    ++l;
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j < n; ++j) {
      if (maxStart[j]) {
        dp[j][i % 2] = max(dp[j - 1][i % 2], dp[maxStart[j] - 1][(i - 1) % 2] +
                                                 j - maxStart[j] + 1);
        continue;
      }
      dp[j][i % 2] = max(dp[j - 1][i % 2], j - maxStart[j] + 1);
    }
  }
  cout << dp[n - 1][k % 2] << endl;
  return 0;
}