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 int minn(long long int a, long long int b) {
  if (a > b) return b;
  return a;
}
long long int maxx2(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int maxx(long long int a, long long int b, long long int c) {
  return maxx2(a, maxx2(b, c));
}
long long int gcdd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcdd(b, a % b);
}
long long int dsum(long long int n) {
  long long int sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
long long int poww(long long int a, long long int b) {
  if (b <= 0) return 1;
  if (b == 1)
    return a;
  else
    return poww(a, b / 2) * poww(a, b / 2) * poww(a, b % 2);
}
string substr1(string s, long long int i, long long int j) {
  string r = s.substr(i, j - i + 1);
  return r;
}
vector<long long int> ans;
long long int visited[100001] = {0};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<long long int> v;
  long long int a;
  for (long long int i = 0; i < n; i++) {
    cin >> a;
    v.push_back(a);
  }
  sort(v.begin(), v.end());
  vector<long long int>::iterator it;
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    it = upper_bound(v.begin(), v.end(), v[i] + 5);
    ans = maxx2(ans, it - v.begin() - i);
  }
  cout << ans << endl;
  return 0;
}