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 long long mod = (long long)1e9 + 7;
const long long inf = (long long)1e16;
const long double eps = 1e-12;
const long long N = (int)1e5 + 5;
const long long LOGN = 19;
const long double PI = 3.14159265358979323846;
inline long long mul(long long a, long long b, long long m = mod) {
  return (long long)(a * b) % m;
}
inline long long add(long long a, long long b, long long m = mod) {
  a += b;
  if (a >= m) a -= m;
  if (a < 0) a += m;
  return a;
}
inline long long power(long long a, long long b, long long m = mod) {
  if (b == 0) return 1;
  if (b == 1) return (a % m);
  long long x = power(a, b / 2, m);
  x = mul(x, x, m);
  if (b % 2) x = mul(x, a, m);
  return x;
}
int n, k;
int aa[5005];
vector<int> a;
unordered_map<int, int> mp;
int dp[5005][5005];
int rec(int pos, int rem) {
  if (pos < 0 || rem <= 0) {
    return 0;
  }
  int &ans = dp[pos][rem];
  if (ans != -1) return ans;
  ans = 0;
  ans = rec(pos - 1, rem);
  int to = mp[a[pos]], i = pos - 1;
  for (i = pos - 1; i >= -1; i--) {
    ans = max(ans, rec(i, rem - 1) + to);
    if (i == -1) break;
    if (a[pos] - a[i] > 5) break;
    to += mp[a[i]];
  }
  ans = max(ans, rec(i, rem - 1) + to);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (0) {
    freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\input.txt", "r",
            stdin);
    freopen("C:\\Users\\Dishant\\Desktop\\Collection-DEV c++\\output.txt", "w",
            stdout);
  }
  int tn;
  cin >> tn >> k;
  for (int i = 0; i < tn; i++) {
    cin >> aa[i];
    if (mp.find(aa[i]) == mp.end()) {
      a.push_back(aa[i]);
    }
    mp[aa[i]]++;
  }
  sort(a.begin(), a.end());
  n = a.size();
  if (k >= n) {
    cout << tn;
    return 0;
  }
  memset(dp, -1, sizeof(dp));
  cout << rec(n - 1, k);
  return 0;
}