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 MAX = 5e3 + 5;
const int OO = 1e9;
int n, nextStart[MAX], memo[MAX][MAX];
int solve(int i, int k) {
  if (k == 0 || i == n) return 0;
  if (memo[i][k] != -1) return memo[i][k];
  return memo[i][k] = max(solve(i + 1, k),
                          solve(nextStart[i], k - 1) + nextStart[i] - i);
}
int main() {
  int k;
  cin >> n >> k;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    v.push_back(a);
  }
  sort(v.begin(), v.end());
  int be = 0;
  for (int i = 0; i < n; i++) {
    while (be < n && v[i] + 5 >= v[be]) be++;
    nextStart[i] = be;
  }
  memset(memo, -1, sizeof memo);
  cout << solve(0, k);
  return 0;
}