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 double pi = 3.141592653589793238462643383279;
const long long INF = 1e18;
void bruteforce() {}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long a[n + 1];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  a[n] = LLONG_MAX;
  sort(a, a + n);
  long long cnt[n];
  for (long long i = 0; i < n; i++) {
    long long curr = 1;
    for (long long j = i + 1; j < n + 1; j++) {
      if (a[j] - a[i] <= 5) {
        curr++;
      } else {
        break;
      }
    }
    cnt[i] = curr;
  }
  long long dp[n + 1][k + 1];
  for (long long i = 0; i < n + 1; i++) {
    for (long long j = 0; j < k + 1; j++) dp[i][j] = 0;
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < k + 1; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (j + 1 <= k)
        dp[i + cnt[i]][j + 1] = max(dp[i + cnt[i]][j + 1], dp[i][j] + cnt[i]);
    }
  }
  cout << dp[n][k];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  while (T--) {
    solve();
  }
  return 0;
}