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;
int n, k;
vector<int> v;
int ind[5005];
int dp[5005][5005];
int pref[5005];
set<int> sett;
map<int, int> mapp;
int sum[5005];
int solve(int it, int ck) {
  if (ck == 0) return 0;
  if (it < 0) return 0;
  if (it + 1 < ck) return 0;
  if (it + 1 == ck) {
    dp[it][ck] = sum[it];
    return dp[it][ck];
  }
  if (dp[it][ck] == -1) {
    if (ck == 1) {
      dp[it][ck] = pref[it];
      return dp[it][ck];
    }
    int ret = solve(it - 1, ck);
    for (int i = ind[it]; i <= it; i++) {
      if (i != 0)
        ret = max(ret, solve(i - 1, ck - 1) + (sum[it] - sum[i - 1]));
      else
        ret = max(ret, solve(i - 1, ck - 1) + (sum[it]));
    }
    dp[it][ck] = ret;
  }
  return dp[it][ck];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    mapp[temp]++;
    v.push_back(temp);
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) sett.insert(v[i]);
  v.clear();
  for (auto e : sett) v.push_back(e);
  n = v.size();
  sum[0] = mapp[v[0]];
  for (int i = 1; i < n; i++) sum[i] = sum[i - 1] + mapp[v[i]];
  for (int i = n - 1; i >= 0; i--) {
    auto a = lower_bound(v.begin(), v.end(), v[i] - 5);
    ind[i] = a - v.begin();
  }
  for (int i = n - 1; i >= 0; i--) {
    if (ind[i] != 0)
      pref[i] = sum[i] - sum[ind[i] - 1];
    else
      pref[i] = sum[i];
  }
  for (int i = 1; i < n; i++) pref[i] = max(pref[i - 1], pref[i]);
  int ans = 0;
  for (int i = k; i >= 1; i--) {
    int tt = solve(n - 1, i);
    ans = max(ans, tt);
  }
  cout << ans << endl;
  return 0;
}