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 MOD = 1000000007;
long long dp[5005][5005];
long long ind[5005];
long long n;
long long recur(long long i, long long k) {
  if (dp[i][k] != -1) return dp[i][k];
  if (k == 0) return dp[i][k] = 0;
  if (i == n) return dp[i][k] = 0;
  return dp[i][k] =
             max(ind[i] - i + 1 + recur(ind[i] + 1, k - 1), recur(i + 1, k));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  for (int i = 0; i < 5005; i++)
    for (int j = 0; j < 5005; j++) dp[i][j] = -1;
  long long k;
  cin >> n >> k;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (long long i = 0; i < n; i++) {
    long long j;
    for (j = i + 1; j < n; j++) {
      if (a[j] - a[i] > 5) break;
    }
    ind[i] = j - 1;
  }
  cout << recur(0, k) << endl;
}