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>
const long long int MOD = 1e9 + 7;
const long long int INF = 1011111111;
const long long int LLINF = 1000111000111000111LL;
const long double EPS = 1e-10;
const long double PI = 3.14159265358979323;
using namespace std;
long long int power(int x, int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return power((x * x), n / 2);
  else
    return x * power((x * x), (n - 1) / 2);
}
long long int dp[5005][5005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, j, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (i = 0; i < n; i++) cin >> v[i];
  sort(v.begin(), v.end());
  vector<pair<pair<long long int, long long int>, long long int> > check;
  vector<long long int>::iterator it;
  for (i = 0; i < n; i++) {
    long long int idx;
    it = upper_bound(v.begin(), v.end(), v[i] + 5);
    if (it == v.end())
      idx = n - 1;
    else {
      idx = it - v.begin();
      idx--;
    }
    check.push_back({{i, idx}, idx - i + 1});
  }
  long long int ans = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j <= k; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (j + 1 <= k)
        dp[i + check[i].second][j + 1] =
            max(dp[i + check[i].second][j + 1], dp[i][j] + check[i].second);
    }
  }
  cout << dp[n][k];
}