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;
int a[5005], Next[5005];
int dp[5005][5005];
int n;
int dfs(int x, int k) {
  if (x == n + 1) return 0;
  if (dp[x][k] != -1) return dp[x][k];
  return dp[x][k] = max(dfs(x + 1, k),
                        (k > 0 ? dfs(Next[x], k - 1) + Next[x] - x : 0));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  memset(dp, -1, sizeof(dp));
  int k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  a[n + 1] = 0x3f3f3f3f;
  int i = 1;
  for (int t = 1; t <= n; t++) {
    while (a[i] - a[t] <= 5) {
      i++;
    }
    Next[t] = i;
  }
  cout << dfs(1, k) << endl;
  cin.get(), cin.get();
  return 0;
}