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 = 5e3 + 7;
int k, it[maxn][maxn], n, a[maxn], f[maxn][maxn], res;
int get(int x, int l) {
  int ans = 0;
  for (; x; x -= (x & -x)) ans = max(ans, it[x][l]);
  return ans;
}
void update(int x, int l) {
  int kk = f[x][l];
  for (; x <= n; x += (x & -x)) it[x][l] = max(it[x][l], kk);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      int kk = lower_bound(a + 1, a + 1 + n, a[i] - 5) - a;
      f[i][j] = get(kk - 1, j - 1) + (i - kk + 1);
      update(i, j);
    }
  res = 0;
  for (int i = 1; i <= n; i++) res = max(res, f[i][k]);
  cout << res;
}