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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
const long long minf = 1e+18;
const long long N = 5010;
const long double ep = 2.71828182845904523536;
using namespace std;
int n, k;
vector<int> xd;
map<int, int> m;
int suma[N];
int siguiente[N];
int used[N][N];
int dp[N][N];
int solve(int l, int ki) {
  if (l >= n) return 0;
  if (ki <= 0) return 0;
  if (used[l][ki]) return dp[l][ki];
  used[l][ki] = true;
  dp[l][ki] = max(solve(l + 1, ki), suma[l] + solve(siguiente[l], ki - 1));
  return dp[l][ki];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    int a;
    cin >> a;
    xd.push_back(a);
    m[a]++;
  }
  sort(xd.begin(), xd.end());
  for (int i = 0; i < n; i++) {
    int u = xd[i];
    int aea = m[u] + m[u + 1] + m[u + 2] + m[u + 3] + m[u + 4] + m[u + 5];
    suma[i] = aea;
    vector<int>::iterator it;
    it = upper_bound(xd.begin(), xd.end(), u + 5);
    siguiente[i] = it - xd.begin();
  }
  cout << solve(0, k) << endl;
  return 0;
}