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;
double eps = 1e-9;
int inf = 1000000005;
long long inff = 1000000000000000005ll;
long long mod = 1000000007;
struct stack {
  vector<long long> s, smin = {LLONG_MAX}, smax = {LLONG_MIN};
  void push(long long x) {
    s.push_back(x);
    smin.push_back(::min(smin.back(), x));
    smax.push_back(::max(smax.back(), x));
  }
  long long pop() {
    long long res = s.back();
    s.pop_back();
    smin.pop_back();
    smax.pop_back();
    return res;
  }
  bool empty() { return s.empty(); }
  long long min() { return smin.back(); }
  long long max() { return smax.back(); }
};
::stack s1, s2;
long long k;
void add(long long x) { s2.push(x); }
void remove() {
  if (s1.empty()) {
    while (!s2.empty()) {
      s1.push(s2.pop());
    }
  }
  s1.pop();
}
bool good() {
  long long mx = max(s1.max(), s2.max());
  long long mn = min(s1.min(), s2.min());
  return mx - mn <= k;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    k = 5;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a.begin(), a.end());
    long long sum = INT_MIN, i = 0;
    for (long long j = 0; j < n; j++) {
      add(a[j]);
      while (!good()) {
        remove();
        i++;
      }
      sum = max(sum, j - i + 1);
    }
    cout << sum << endl;
  }
  return 0;
}