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 long long INF = 1e18;
const double eps = 1e-9;
long long n, k, dp[5005][5005], sum[5005], a[5005], ans;
int main() {
  while (~scanf("%lld %lld", &n, &k)) {
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
    }
    sort(a + 1, a + 1 + n);
    for (long long i = 1; i <= n; i++) {
      long long j;
      for (j = i; j <= n; j++) {
        if (a[j] - a[i] > 5) {
          break;
        }
      }
      sum[i] = j - 1;
    }
    dp[0][0] = 0;
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j <= k; j++) {
        dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
        ans = max(ans, dp[i + 1][j]);
        if (j != k) {
          dp[sum[i + 1]][j + 1] =
              max(dp[sum[i + 1]][j + 1], dp[i][j] + sum[i + 1] - (i + 1) + 1);
          ans = max(ans, dp[sum[i + 1]][j + 1]);
        }
      }
    }
    printf("%lld\n", ans);
  }
}