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>
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
using namespace std;
int cases, ct = 0;
int main() {
  std::ios::sync_with_stdio(false);
  ;
  long long n, k;
  vector<long long> A;
  while (cin >> n) {
    A.clear();
    long long i, j;
    for (i = 0; i < n; i++) {
      cin >> j;
      A.push_back(j);
    }
    sort(A.begin(), A.end());
    long long res = 1, cnt = 1;
    long long now_diff = 0, shuru = 0, fl = 0;
    for (i = 0; i < A.size(); i++) {
      while (A[shuru] - A[i] <= 5) {
        if (shuru >= n) break;
        shuru++;
      }
      res = max(res, shuru - i);
    }
    cout << res << endl;
  }
}