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;
long long mod = 1000000007;
long long pow(long long x, long long y) {
  if (y == 0) return 1;
  long long z = pow(x, y / 2);
  if (y % 2) return z * z * x;
  return z * z;
}
long long fpow(long long a, long long b) {
  long long p = a, ans = 1;
  while (b) {
    if (b & 1) ans = (ans * p) % mod;
    p = (p * p) % mod;
    b >>= 1;
  }
  return ans;
}
long long mpow(long long x, long long y) {
  if (y == 0) return 1;
  long long z = mpow(x, y / 2) % mod;
  if (y % 2) return (((z * z) % mod) * x) % mod;
  return (z * z) % mod;
}
long long mod_inv(long long x) { return mpow(x, mod - 2) % mod; }
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long gcm(long long x, long long y) { return (x * y) / (gcd(x, y)); }
long long no_factor(long long x) {
  long long i, count = 0;
  for (i = 1; i * i <= x; i++)
    if (x % i == 0 && i * i != x)
      count += 2;
    else if (x % i == 0)
      count++;
  return count;
}
long long a[200001], n;
long long bs(long long k) {
  long long ans = 0, s = 0, e = n - 1, mid;
  while (s <= e) {
    mid = s + (e - s) / 2;
    if (a[mid] <= k) {
      ans = max(mid, ans);
      s = mid + 1;
    } else
      e = mid - 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long maxi = 0, curr;
  for (i = 0; i < n; i++) {
    curr = bs(a[i] + 5);
    maxi = max(maxi, curr - i + 1);
  }
  cout << maxi;
  return 0;
}