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 N = 3e5 + 123;
const int mod = 1e9 + 7;
const int inf = 1e9 + 123;
const long long INF = 1e18 + 123;
const double eps = (double)(1e-9);
const double pi = 3.14159265358979323846264;
const int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
const int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
const pair<int, int> dir[] = {make_pair(-1, 0), make_pair(0, -1),
                              make_pair(1, 0), make_pair(0, 1)};
const pair<int, int> dirr[] = {
    make_pair(-2, 1), make_pair(1, -2), make_pair(2, -1),  make_pair(-1, 2),
    make_pair(2, 1),  make_pair(1, 2),  make_pair(-2, -1), make_pair(-1, -2)};
int n, a[N];
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  int j = 1, mx = 0;
  for (int i = 1; i <= n; i++) {
    while (j <= n && a[j] - a[i] <= 5) {
      j++;
      mx = max(mx, j - i);
    }
  }
  cout << mx;
  return 0;
}