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;
template <typename T1, typename T2>
inline void chmin(T1& a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1& a, T2 b) {
  if (a < b) a = b;
}
const int MOD = 998244353;
const int MAX = 200060;
const double pi = acos(-1);
const double EPS = 1e-12;
const long long INF = 1e18;
int dp[5050][5050];
bool used[5050][5050];
int N;
vector<int> v;
int func(int i, int rem) {
  if (used[i][rem]) return dp[i][rem];
  int ret = 0;
  if (rem == 0) {
    used[i][rem] = true;
    dp[i][rem] = 0;
    return 0;
  } else if (i >= N) {
    used[i][rem] = true;
    dp[i][rem] = 0;
    return 0;
  } else {
    int ret1 = 0, ret2 = 0;
    auto itr = upper_bound(v.begin(), v.end(), v[i] + 5);
    int id = itr - v.begin();
    ret1 = (id - i) + func(id, rem - 1);
    ret2 = func(i + 1, rem);
    int ret = max(ret1, ret2);
    used[i][rem] = true;
    dp[i][rem] = ret;
    return ret;
  }
}
void solve() {
  cin >> N;
  int K;
  cin >> K;
  for (int i = 0; i < N; ++i) {
    int x;
    cin >> x;
    v.push_back(x);
  }
  v.push_back(2e9);
  sort(v.begin(), v.end());
  ;
  long long ans = func(0, K);
  cout << (ans) << '\n';
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
}