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;
vector<int> A;
int binsearch(int l, int r, int a) {
  while (r - l > 1) {
    int midd = (r + l) / 2;
    if (A[midd] <= a) {
      l = midd;
    } else {
      r = midd;
    }
  }
  return r - 1;
}
int main() {
  int n, k;
  cin >> n >> k;
  A.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
  sort(A.begin(), A.end());
  vector<vector<int> > dp(n + 1, vector<int>(k + 1, 0));
  int max1 = 0;
  for (int j = 0; j < k; j++) {
    for (int i = 0; i < n; i++) {
      int y = binsearch(0, n, A[i] + 5);
      if (i != 0) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      }
      dp[y + 1][j + 1] = max(dp[y + 1][j + 1], dp[i][j] + y - i + 1);
      max1 = max(max1, dp[y + 1][j + 1]);
    }
  }
  cout << max1;
  return 0;
}