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;
ifstream in("input.txt");
const long long mod = 1e9 + 7;
const int N = 1e5 + 5;
const long double eps = 1e-7;
long long mod_pow(long long a, long long b) {
  if (b == 0) return 1LL;
  long long res = mod_pow(a, b >> 1);
  res = (res * res) % mod;
  if (b & 1LL) res = (res * a) % mod;
  return res;
}
vector<char> seive(long long n) {
  vector<char> is_prime(n + 1, true);
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i * i <= n; i++) {
    if (is_prime[i]) {
      for (int j = i * i; j <= n; j += i) is_prime[j] = false;
    }
  }
  return is_prime;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long v[n];
  for (int(i) = (0); (i) < (n); (i)++) {
    cin >> v[i];
  }
  sort(v, v + n);
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    long long pt = (lower_bound(v, v + n, v[i] - 5) - v);
    ans = max(ans, i - pt + 1);
  }
  cout << ans << "\n";
  return 0;
}