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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
void print(long long int* a, int n) {
  int i;
  for (i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
int arrMax(int* a, int n) {
  int i, max = a[0];
  for (i = 1; i < n; i++) {
    if (a[i] > max) max = a[i];
  }
  return max;
}
void print2d(int a[][100], int m, int n) {
  int i, j;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      cout << setw(2) << a[i][j];
    }
    cout << endl;
  }
}
void fio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
int n;
int* a;
bool isPossible(int mid) {
  int i = 0;
  while (i + mid <= n) {
    int m = a[i + mid - 1] - a[i];
    if (m <= 5) return 1;
    i++;
  }
  return 0;
}
int main() {
  fio();
  cin >> n;
  a = new int[n];
  int i;
  for (i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int e = n, s = 1;
  int mid, ans;
  while (e >= s) {
    mid = (e + s) / 2;
    if (isPossible(mid))
      ans = mid, s = mid + 1;
    else
      e = mid - 1;
  }
  cout << ans << endl;
}