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 double PI = atan(1.0) * 4;
const int INF = 0x7ffffff;
const long long MOD = 1000000007;
const int maxn = 5010;
long long GCD(long long a, long long b) { return b == 0 ? a : GCD(b, a % b); }
int a[maxn];
int cnt[maxn];
int dp[maxn][maxn];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    while (i + cnt[i] <= n && a[i + cnt[i]] - a[i] <= 5) cnt[i]++;
  }
  for (int i = 1; i <= n; i++) {
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++) {
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      dp[i + cnt[i] - 1][j] =
          max(dp[i + cnt[i] - 1][j], dp[i - 1][j - 1] + cnt[i]);
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
    }
  }
  printf("%d\n", dp[n][k]);
  return 0;
}