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 long long mod = 1000000007;
vector<int> a;
vector<vector<int> > dp;
int solve(int at, int k) {
  if (at <= 0 or k == 0) return 0;
  if (dp[at][k] == -1) {
    int x = upper_bound(a.begin(), a.end(), a[at - 1] - 6) - a.begin();
    dp[at][k] = max(solve(at - 1, k), solve(x, k - 1) + at - x);
  }
  return dp[at][k];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a.begin(), a.end());
  dp.assign(n + 1, vector<int>(k + 1, -1));
  cout << solve(n, k) << "\n";
  return 0;
}