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;
const int maxn = 5005;
inline void read(int &x) {
  x = 0;
  int f = 1;
  char ch = 0;
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  x *= f;
}
int n, k, a[maxn], dp[maxn][maxn], lp[maxn], ans = 1, mx[maxn][maxn];
int main() {
  read(n);
  read(k);
  for (register int i = 1; i <= n; ++i) read(a[i]);
  sort(a + 1, a + n + 1);
  for (register int i = 1; i <= n; ++i) {
    for (register int j = 1; j <= i; ++j) {
      lp[i] = j;
      if (a[i] - a[j] <= 5) break;
    }
  }
  for (register int i = 1; i <= n; ++i) {
    for (register int j = 1; j <= k; ++j) {
      dp[i][j] = mx[lp[i] - 1][j - 1] + i - lp[i] + 1;
      mx[i][j] = max(dp[i][j], mx[i - 1][j]);
      ans = max(ans, dp[i][j]);
    }
  }
  cout << ans;
  return 0;
}