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, k, a[200010], memo[5010][5010];
multiset<pair<long long, long long> > s;
long long dp(long long obj, long long falta) {
  if (memo[obj][falta] != -1) return memo[obj][falta];
  if (falta == 0) return memo[obj][falta] = 0;
  if (obj > n) return memo[obj][falta] = 0;
  long long resp = -9123456789;
  auto it = s.upper_bound({a[obj] + 5, 9123456789});
  if (it == s.end()) {
    resp = n - obj + 1;
    return memo[obj][falta] = resp;
  }
  it--;
  pair<long long, long long> aux = *it;
  resp = max(resp, dp(aux.second + 1, falta - 1) + aux.second - obj + 1);
  resp = max(resp, dp(obj + 1, falta));
  return memo[obj][falta] = resp;
}
void solve() {
  cin >> n >> k;
  memset(memo, -1, sizeof(memo));
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n; i++) s.insert({a[i], i});
  cout << dp(1, k) << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}