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 = 5000 + 10;
long long n, k, ans, num[maxn], pre[maxn], dp[maxn][maxn];
inline long long min(long long a, long long b) { return a < b ? a : b; }
inline long long max(long long a, long long b) { return a > b ? a : b; }
void init() {
  sort(num + 1, num + n + 1);
  for (int i = n, j = i - 1; i >= 1; i--) {
    for (; j >= 1; j--) {
      if (num[i] - num[j] > 5) {
        pre[i] = i - j;
        break;
      }
    }
    if (j < 1) pre[i] = i - j;
  }
  dp[1][1] = 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> num[i];
  init();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= min(i, k); j++)
      dp[i][j] = max(dp[i - 1][j], dp[max(i - pre[i], j - 1)][j - 1] +
                                       min(pre[i], i - j + 1));
  cout << dp[n][k] << endl;
}