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 = int(5e3 + 10);
int arr[maxn];
int pos[maxn];
int dp[maxn][maxn];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", arr + i);
  }
  sort(arr + 1, arr + 1 + n);
  if (n == k || arr[n] - arr[1] <= 5) {
    printf("%d\n", n);
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    int p = lower_bound(arr + 1, arr + 1 + i, arr[i] - 5) - arr;
    pos[i] = p;
  }
  for (int i = 1; i <= n; ++i) {
    int ed = min(i, k);
    for (int j = 1; j <= ed; ++j) {
      dp[i][j] = max(dp[i][j - 1],
                     max(i - pos[i] + 1 + dp[pos[i] - 1][j - 1], dp[i - 1][j]));
    }
  }
  printf("%d\n", dp[n][k]);
}