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 N = 2e5 + 5;
int t, id, n, m, x, y, k, c, p, dif, ans, sum, pre, rem, cur, tmp, tot, r, l, u,
    d, xx, yy;
int a[N], vis[N], f[N], b[N];
bool fl, ok;
vector<int> adj[N], v;
int dp[5005][5005];
int sol() {
  for (int i = (int)n; i >= 0; --i) {
    for (int rem = 0; rem <= (int)k; ++rem) {
      dp[i][rem] = 0;
      if (i == n || !rem) continue;
      dp[i][rem] = dp[i + 1][rem];
      dp[i][rem] = max(dp[i][rem], b[i] + dp[i + b[i]][rem - 1]);
    }
  }
  return dp[0][k];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < (int)n; ++i) scanf("%d", &a[i]);
  sort(a, a + n);
  for (int i = 0; i < (int)n; ++i) {
    b[i] = upper_bound(a, a + n, a[i] + 5) - a - i;
  }
  printf("%d\n", sol());
  return 0;
}