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 form k teams for yet another new programming competition. As you know, the more students are involved in competition the more probable the victory of your university is! So you have to form no more than k (and at least one) non-empty teams so that the total number of students in them is maximized. But you also know that each team should be balanced. It means that the programming skill of each pair of students in each team should differ by no more than 5. Teams are independent from one another (it means that the difference between programming skills of two students from two different teams does not matter).

It is possible that some students not be included in any team at all.

Your task is to report the maximum possible total number of students in no more than k (and at least one) non-empty balanced teams.

If you are Python programmer, consider using PyPy instead of Python when you submit your code.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 5000) — the number of students and the maximum number of teams, correspondingly.

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 total number of students in no more than k (and at least one) non-empty balanced teams.

Examples

Input


5 2
1 2 15 15 15


Output


5


Input


6 1
36 4 1 25 9 16


Output


2


Input


4 4
1 10 100 1000


Output


4
#include <bits/stdc++.h>
using namespace std;
const long long N = 5005;
long long dp[N][N];
long long Prev[N];
long long a[N];
long long solve(long long idx, long long team) {
  if (idx < 0) return -1e9;
  if (team == 0) return 0;
  long long &ans = dp[idx][team];
  if (ans != -1) return ans;
  ans = solve(idx - 1, team);
  ans = max(ans, (idx - Prev[idx]) + solve(Prev[idx], team - 1));
  return ans;
}
signed main() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  long long l = 1;
  for (long long r = 1; r <= n; r++) {
    while (a[r] - a[l] > 5) l++;
    Prev[r] = l - 1;
  }
  memset(dp, -1, sizeof dp);
  cout << solve(n, k);
  return 0;
}
