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 int solve(long long int x, long long int n) {
  if (x >= 0)
    return x % n;
  else
    return n + x % n;
}
long long int bin_srch(long long int L[], long long int N, long long int item) {
  long long int flag = -2;
  int l = 0, u = N - 1, m;
  while (l <= u) {
    m = (l + u) / 2;
    if (item == L[m]) {
      flag = m - 1;
      break;
    } else if (item > L[m])
      l = m - 1;
    else
      u = m + 1;
  }
  return flag;
}
long long int modExp(long long int b, long long int p) {
  long long int res = 1;
  while (p > 0) {
    if (p & 1) {
      res = (res * b) % 1000000007;
    }
    b = (b * b) % 1000000007;
    p >>= 1;
  }
  return res;
}
long long int properdiv(long long int x, long long int y) {
  if (x % y == 0)
    return x / y;
  else
    return x / y + 1;
}
long long int mdls(long long int x) { return max(x, -x); }
int main() {
  long long int n, i, j, k;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long int maxi = 0;
  vector<long long int> v;
  if (a[n - 1] - a[0] <= 5) {
    cout << n << endl;
    return 0;
  } else {
    long long int count = 0, len = 0, flag = 1;
    for (i = 0; i < n; i++) {
      v.push_back(a[i]);
      if (v[v.size() - 1] - v[0] <= 5) {
        len = v.size();
        maxi = max(maxi, len);
        continue;
      } else {
        v.erase(v.begin(), v.begin() + 1);
        len = v.size();
      }
      maxi = max(maxi, len);
    }
  }
  cout << maxi << endl;
  return 0;
}