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;
bool pr[1000007];
long long fpow(long long x, long long y) {
  x = x % 1000000007;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= 1000000007;
    y = y >> 1;
    x = x * x;
    x %= 1000000007;
  }
  return res;
}
long long inv(long long a, long long m = 1000000007) {
  long long c = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += c;
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long cnt[n];
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < n; i++) {
    while (i + cnt[i] < n && a[i + cnt[i]] - a[i] <= 5) cnt[i]++;
  }
  long long dp[n + 1][k + 2];
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i + cnt[i]][j + 1] = max(dp[i + cnt[i]][j + 1], dp[i][j] + cnt[i]);
    }
  }
  cout << dp[n][k];
}