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;
const int nax = 5005;
int dp[nax][nax][2];
int a[5005];
int pref[5005];
map<int, int> m;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int n, k;
  int t[5005];
  cin >> n >> k;
  for (int i = 1; n >= i; ++i) cin >> t[i], m[t[i]]++;
  sort(t + 1, t + n + 1);
  t[0] = -2;
  int ile = 1;
  for (int i = 1; n >= i; ++i) {
    while (t[i - 1] == t[i]) i++;
    if (i <= n) {
      a[ile] = t[i];
      ile++;
    }
  }
  ile--;
  for (int i = 1; ile >= i; ++i) pref[i] = pref[i - 1] + m[a[i]];
  dp[1][1][1] = m[a[1]];
  for (int i = 1; ile >= i; ++i) {
    for (int mas = 0; k >= mas; mas++) {
      for (int j = i - 1; i - 6 < j; --j) {
        if (j <= 0) continue;
        if (a[i] - a[j] <= 5 and mas >= 1)
          dp[i][mas][0] = max(dp[i][mas][0], dp[j][mas][1] + pref[i] - pref[j]);
        dp[i][mas][0] = max(dp[i][mas][0], max(dp[j][mas][0], dp[j][mas][1]));
        if (mas >= 1)
          dp[i][mas][1] =
              max(dp[i][mas][1],
                  max(dp[j][mas - 1][0], dp[j][mas - 1][1]) + m[a[i]]);
      }
    }
  }
  int ans = 0;
  for (int i = 1; k >= i; ++i)
    ans = max(ans, max(dp[ile][i][0], dp[ile][i][1]));
  cout << ans;
  return 0;
}