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 N = 5e3 + 5;
const int inf = 6969696;
int dp[N][N];
int a[N];
int e[N];
int n, k;
int go(int pos, int left) {
  if (left < 0) return -inf;
  if (pos == n) return 0;
  int &cur = dp[pos][left];
  if (cur != -1) return cur;
  cur = go(pos + 1, left);
  cur = max(cur, e[pos] - pos + go(e[pos], left - 1));
  return cur;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  int i = 0, ptr = 0;
  while (i < n) {
    while (ptr < n and a[ptr] <= a[i] + 5) ptr++;
    e[i] = ptr;
    i++;
  }
  memset(dp, -1, sizeof dp);
  printf("%d\n", go(0, k));
  return 0;
}