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 MOD = 1000000007;
const int MAX = 1000005;
int SetBit(int n, int x) { return n | (1 << x); }
int ClearBit(int n, int x) { return n & ~(1 << x); }
int ToggleBit(int n, int x) { return n ^ (1 << x); }
bool CheckBit(int n, int x) { return (bool)(n & (1 << x)); }
int arr[200005];
int main(void) {
  int n, i, p, ans;
  while (scanf("%d", &n) != EOF) {
    for (i = 0; i < n; i++) scanf("%d", &arr[i]);
    sort(arr, arr + n);
    ans = 0;
    for (i = 0; i < n; i++) {
      p = upper_bound(arr, arr + n, arr[i] + 5) - arr;
      if (p == n) --p;
      if (arr[i] + 5 < arr[p]) --p;
      ans = max(ans, p - i + 1);
    }
    printf("%d\n", ans);
  }
  return 0;
}