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 dp[5001][5001];
int cnt[5001];
int n, k;
int main() {
  for (int i = 0; i < 5001; i++)
    for (int ii = 0; ii < 5001; ii++) dp[i][ii] = 0;
  cin >> n >> k;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int tmp;
    cin >> tmp;
    v.push_back(tmp);
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    int pos = upper_bound(v.begin(), v.end(), v[i] + 5) - v.begin() - 1;
    cnt[i] = pos - i + 1;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (j + 1 <= k) {
        dp[i + cnt[i]][j + 1] = max(dp[i + cnt[i]][j + 1], dp[i][j] + cnt[i]);
      }
    }
  }
  cout << dp[n][k] << endl;
  return 0;
}