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;
void DBG() { cerr << "]\n"; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << h;
  if (sizeof...(t)) {
    cerr << ", ";
  }
  DBG(t...);
}
int n, m, ans = -1;
int a[5005], dp[5005][5005];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  dp[1][1] = 1;
  for (int i = 1; i <= n; i++) {
    int la = i;
    while (abs(a[la] - a[i]) <= 5 && la >= 1) la--;
    for (int j = 1; j <= m; j++)
      dp[i][j] = max(dp[i - 1][j], dp[la][j - 1] + (i - la));
  }
  for (int i = 1; i <= m; i++) ans = max(ans, dp[n][i]);
  cout << ans << endl;
  return 0;
}