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;
const int MAXN = 2e5 + 1;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int EPS = 1e-6;
long long a[MAXN], b[MAXN];
vector<int> g[MAXN], v;
bool used[MAXN];
void freopen() {
  freopen("A.in", "r", stdin);
  freopen("A.out", "w", stdout);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long n, m, k, cnt = 0, sum = 0, mx = 0, mn = INF;
  bool y = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + 1 + n);
  long long j = 1;
  for (int i = 1; i <= n; i++) {
    while (a[j] - a[i] <= 5 && j <= n) {
      j++;
    }
    j--;
    mx = max(j - i, mx);
  }
  cout << mx + 1;
  return 0;
}