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;
vector<long long> vprime;
void SieveOfEratosthenes(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) vprime.push_back(p);
}
int countFreq(string &pat, string &txt) {
  int M = pat.length();
  int N = txt.length();
  int res = 0;
  for (int i = 0; i <= N - M; i++) {
    int j;
    for (j = 0; j < M; j++)
      if (txt[i + j] != pat[j]) break;
    if (j == M) {
      res++;
      j = 0;
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long a, b, c, d, e, g, h, k, l, m, n, t, f,
      f1 = 0, flag2 = 0, flag = 0, flag1 = 0, co = 0, co1 = 0, co2 = 0, sum = 0,
      sum1 = 0, ma = 0, ma1 = 0, ma2 = 0, mi = 100000000000, mi1 = 100000000000,
      mi2 = 100, mi3 = 1000;
  long long co3 = 0, co4 = 0, co5 = 0, co6 = 0, co7 = 0, co8 = 0, mul = 1,
            sum2 = 0;
  long long arr[100001] = {0}, arr1[100001] = {0};
  long long a1, a2, b1, b2, c1, c2, a3, a4, b3, b4, b5, b6, m1, m2, k1, l1, m3,
      m4, d1, d2;
  double pi = 2 * acos(0.0);
  char ch;
  string str, str1, str2, str3, str4, str5;
  set<long long> s1;
  set<long long> s2;
  set<long long> s3;
  stack<long long> st;
  int x[] = {1, 0, 0, -1};
  int y[] = {0, 1, -1, 0};
  cin >> n;
  vector<long long> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  sort(v.begin(), v.end());
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  k = 1;
  co = 1;
  for (int i = 0; i < n; i++) {
    for (int j = k; j < n; j++) {
      if ((v[j] - v[i]) <= 5) {
        co++;
      } else {
        ma = max(ma, co);
        co--;
        k = j;
        break;
      }
    }
    ma = max(ma, co);
    if (ma >= (n - i)) break;
  }
  cout << ma << endl;
  return 0;
}