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;
void init_code() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long dp[5002][5002];
long long solve(vector<long long>& v, long long k, long long n, long long idx) {
  if (k == 0) return 0;
  if (idx >= n) return 0;
  if (dp[idx][k] != -1) return dp[idx][k];
  auto i = (upper_bound(v.begin(), v.end(), v[idx] + 5) - v.begin());
  long long cnt = i - idx;
  long long temp = solve(v, k - 1, n, i) + cnt;
  temp = max(temp, solve(v, k, n, idx + 1));
  dp[idx][k] = temp;
  return temp;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  sort(v.begin(), v.end());
  memset(dp, -1, sizeof(dp));
  cout << solve(v, k, n, 0);
}
signed main() {
  init_code();
  long long tc = 1;
  long long num = 1;
  while (tc--) {
    num++;
    solve();
  }
}