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;
template <typename T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << '[';
  int b = 0;
  for (const auto& a : v) o << (b++ ? ", " : "") << a;
  o << ']';
  return o;
}
template <typename T, typename U>
ostream& operator<<(ostream& o, const pair<T, U>& p) {
  o << '(' << p.first << ", " << p.second << ')';
  return o;
}
int n, k;
vector<int> A;
int dp[5050][5050];
int f(int i, int t) {
  if (i == n or t == k) return 0;
  int& x = dp[i][t];
  if (x != -1) return x;
  int j = upper_bound(begin(A), end(A), A[i] + 5) - begin(A);
  x = max(f(i + 1, t), f(j, t + 1) + j - i);
  return x;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  A.assign(n, 0);
  for (auto& a : A) cin >> a;
  sort(begin(A), end(A));
  for (int i = 0; i <= n; i++) fill(dp[i], dp[i] + k + 1, -1);
  cout << f(0, 0) << endl;
}