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 SuperINF = 1e9;
const long long MAX_SIZE = 1e3 + 4, INF = 10, MOD = 998244353;
const double eps = 1e-6;
double pi = 3.14159265358979323846;
void files() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
vector<long long> arr;
int n;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long g;
    cin >> g;
    arr.push_back(g);
  }
  sort(arr.begin(), arr.end());
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    auto g = upper_bound(arr.begin(), arr.end(), arr[i] + 5);
    ans = max(ans, (long long)(g - (arr.begin() + i)));
  }
  cout << ans;
  return 0;
}