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 maxn = 5100;
int n, k;
int dp[maxn][maxn];
int num[maxn];
int mx[maxn][maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
  sort(num + 1, num + n + 1);
  memset(dp, -0x3f, sizeof(dp));
  dp[1][1] = 1;
  mx[1][1] = 1;
  dp[1][0] = 0;
  mx[1][0] = 0;
  int p = 1;
  int ans = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j <= min(i, k); j++) {
      if (j == 0) {
        dp[i][j] = 0;
        mx[i][j] = 0;
        continue;
      }
      while (num[i] - num[p] > 5) p++;
      dp[i][j] = mx[p - 1][j - 1] + (i - p + 1);
      mx[i][j] = max(dp[i][j], mx[i - 1][j]);
      ans = max(ans, dp[i][j]);
    }
  }
  printf("%d\n", ans);
  return 0;
}