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 A[500005];
long long n;
bool ok(long long len) {
  long long i = 1;
  long long j = len;
  for (; j <= n; j++, i++) {
    if ((A[j] - A[i]) < 6) {
      return true;
    }
  }
  return false;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt = 1;
  while (tt--) {
    cin >> n;
    for (long long i = 1; i <= n; i++) cin >> A[i];
    long long l = 1, r = n;
    sort(A + 1, A + n + 1);
    long long Ans = 1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (ok(mid)) {
        l = mid + 1;
        Ans = max(Ans, mid);
      } else
        r = mid - 1;
    }
    cout << Ans << '\n';
  }
  return 0;
}