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;
using ii = pair<long long, long long>;
using vb = vector<bool>;
using vi = vector<long long>;
using vii = vector<ii>;
using vvi = vector<vi>;
using vvii = vector<vii>;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long k;
  cin >> k;
  vi a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort((a).begin(), (a).end());
  vvi dp(n + 1, vi(k + 1, 0));
  for (long long i = n - 1; i >= 0; i--) {
    for (long long team = 1; team <= k; team++) {
      long long req = a[i] + 5;
      long long j = upper_bound((a).begin(), (a).end(), req) - begin(a);
      dp[i][team] = max(dp[i + 1][team], (j - i) + dp[j][team - 1]);
    }
  }
  cout << dp[0][k] << '\n';
}