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 n, k, a[N], last[N], sum[N], cnt[N], num[N], lastsum[N], dp[N][N];
int main() {
  scanf("%d%d", &n, &k);
  for (register int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + n);
  int idx = 0;
  for (register int i = 1; i <= n; ++i) {
    if (a[i] != a[i - 1]) {
      num[++idx] = a[i];
    }
    cnt[idx]++;
  }
  n = idx;
  for (register int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + cnt[i];
  int j = 1;
  for (register int i = 1; i <= n; ++i) {
    while (j < i && num[i] - num[j] > 5) j++;
    last[i] = j - 1;
    lastsum[i] = sum[i] - sum[j - 1];
  }
  for (register int i = 1; i <= n; ++i) {
    for (register int j = 1; j <= k; ++j) {
      dp[i][j] = max(dp[i - 1][j], dp[last[i]][j - 1] + lastsum[i]);
    }
  }
  cout << dp[n][k];
  return 0;
}