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 N = 5005;
int p[N];
int dp[N][N];
struct node {
  int left, right, w;
} g[N];
int main() {
  int n, k;
  cin >> n >> k;
  g[1].right = g[1].left = g[1].w = 1;
  for (int i = 1; i <= n; i++) cin >> p[i];
  sort(p + 1, p + n + 1);
  for (int i = 2; i <= n; i++) {
    for (int j = g[i - 1].left; j <= i; j++) {
      if (p[i] - p[j] <= 5) {
        g[i].left = j;
        break;
      }
    }
    g[i].right = i;
    g[i].w = i - g[i].left + 1;
  }
  for (int i = 0; i <= k; i++) dp[0][i] = 0;
  for (int i = 0; i <= n; i++) dp[i][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      if (i - g[i].w >= 0)
        dp[i][j] = max(dp[i - g[i].w][j - 1] + g[i].w, dp[i - 1][j]);
      else
        dp[i][j] = dp[i - 1][j];
    }
  }
  cout << dp[n][k] << endl;
}