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;
char ar[140000];
vector<long long> SieveOfEratosthenes(long long n) {
  vector<long long> result;
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  int count = 1;
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) {
        prime[i] = false;
      }
      count++;
    }
  }
  for (long long p = 2; p <= n; p++)
    if (prime[p]) {
      result.push_back(p);
    }
  return result;
}
vector<long long> Divisors(long long n) {
  vector<long long> res;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        res.push_back(i);
      } else {
        res.push_back(i);
        res.push_back(n / i);
      }
    }
  }
  return res;
}
long long Div(long long n) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return i;
  }
  return n;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<long long> arr(n, 0);
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    sort(arr.begin(), arr.end());
    int count = 1;
    int ans = 0;
    int j = 0;
    for (int i = 0; i < n; ++i) {
      while (j < n && arr[j] - arr[i] <= 5) {
        ++j;
        ans = max(ans, j - i);
      }
    }
    cerr << "The ans is: ";
    cout << ans << endl;
  }
  return 0;
}