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;
long long n, d, a[100069];
pair<long long, long long> dp[100069];
int main() {
  long long i, j, sm, c, lh, rh, md, zz = -1, zm, z;
  scanf("%lld%lld", &n, &d);
  for (i = 1; i <= n; i++) {
    scanf("%lld", a + i);
  }
  sort(a + 1, a + n + 1);
  for (lh = 0, rh = n; lh <= rh;) {
    md = (lh + rh) / 2;
    for (j = 0, i = 1; i <= n; i++) {
      for (; j < n && a[j + 1] < a[i] - 5; j++)
        ;
      dp[i] = max(dp[i - 1], {dp[j].first + i - j - md, dp[j].second + 1});
    }
    sm = dp[n].first;
    c = dp[n].second;
    if (c >= d) {
      zz = md;
      zm = sm;
      lh = md + 1;
    } else {
      rh = md - 1;
    }
  }
  if (zz != -1) {
    z = zm + zz * d;
  } else {
    z = n;
  }
  printf("%lld\n", z);
}