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 N = 1e5 + 10;
const long long M = 1e7;
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const long long inf = -1e18;
bool isprime(long long x) {
  if (x == 1) return false;
  long long y = sqrt(x);
  for (int i = 2; i <= y; i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
long long bin_pow(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (a == 0) return 0;
  if (a == 1) return a;
  if (b % 2 == 0) {
    long long pw = bin_pow(a, b / 2);
    return pw * pw;
  } else {
    return bin_pow(a, b - 1) * a;
  }
}
long long n, k, ans, a[5010], us[5010];
long long d[5001][5001];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  long long l = 1;
  for (int i = 1; i <= n; i++) {
    while (a[i] - a[l] > 5) l++;
    us[i] = l;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      d[i][j] = max(d[i][j], (i - us[i] + 1) + d[us[i] - 1][j - 1]);
      d[i][j] = max(d[i][j], d[i - 1][j]);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      ans = max(ans, d[i][j]);
    }
  }
  cout << ans;
}