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>
namespace io {
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
char F[200];
inline void write(long long x) {
  if (x == 0) return (void)(putchar('0'));
  long long tmp = x > 0 ? x : -x;
  if (x < 0) putchar('-');
  int cnt = 0;
  while (tmp > 0) {
    F[cnt++] = tmp % 10 + '0';
    tmp /= 10;
  }
  while (cnt > 0) putchar(F[--cnt]);
}
}  // namespace io
using namespace io;
using namespace std;
int vis[5100], n, k;
long long a[5100], f[5100][5100];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + n + 1);
  for (int j = 1; j <= k; ++j) {
    for (int l = 1, i = 1; i <= n; ++i) {
      f[i][j] = f[i - 1][j];
      while (l < i && a[i] - a[l] > 5) ++l;
      f[i][j] = max(f[i][j], f[l - 1][j - 1] + (i - l + 1));
    }
  }
  printf("%lld\n", f[n][k]);
}