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 = 5e3 + 10;
long long mem[N][N], n, k, a[N], oo = 1e9 + 10;
long long dp(int pos, int k) {
  if (k < 0) {
    return -oo;
  }
  if (pos == n) {
    return 0;
  }
  long long &ret = mem[pos][k];
  if (ret != -1) {
    return ret;
  }
  ret = dp(pos + 1, k);
  long long d = upper_bound(a + pos, a + n, a[pos] + 5) - a;
  ret = max(ret, d - pos + dp(d, k - 1));
  return ret;
}
int main() {
  memset(mem, -1, sizeof mem);
  scanf("%lld", &n);
  scanf("%lld", &k);
  for (int i = 0; i < n + 0; i++) {
    scanf("%lld", &a[i]);
  };
  sort(a, a + n);
  long long ans = dp(0, k);
  printf("%lld\n", ans);
}