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 = 2e5 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
int n, k, a[N];
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    sort(a + 1, a + 1 + n);
    priority_queue<int, vector<int>, greater<int> > q;
    while (!q.empty()) q.pop();
    q.push(a[1]);
    int ans = 0;
    for (int i = 2; i <= n; i++) {
      int t = q.top(), ts = q.size();
      if (a[i] > t + 5) ans = max(ans, ts);
      while (a[i] > t + 5) {
        q.pop();
        if (!q.empty())
          t = q.top();
        else
          break;
      }
      q.push(a[i]);
    }
    int t = q.size();
    ans = max(ans, t);
    printf("%d\n", ans);
  }
  return 0;
}