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 = 1e6 + 1;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int add(long long int x, long long int y) {
  x += y;
  while (x >= 1000000007) x -= 1000000007;
  while (x < 0) x += 1000000007;
  return x;
}
long long int mul(long long int x, long long int y) {
  return (x * 1ll * y) % 1000000007;
}
long long int binpow(long long int x, long long int y) {
  long long int z = 1;
  while (y) {
    if (y & 1) {
      z = mul(z, x);
    }
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
long long int inv(long long int x) { return binpow(x, 1000000007 - 2); }
long long int divide(long long int x, long long int y) {
  return mul(x, inv(y));
}
long long int fact[N];
void precalc() {
  fact[0] = 1;
  for (long long int i = 1; i < N; i++) {
    fact[i] = mul(fact[i - 1], i);
  }
}
long long int C(long long int n, long long int k) {
  return divide(fact[n], mul(fact[k], fact[n - k]));
}
bool comp(pair<long long int, long long int> &a,
          pair<long long int, long long int> &b) {
  if (a.second > b.second) {
    return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int tc;
  tc = 1;
  while (tc--) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a.begin(), a.end());
    long long int ans = 1;
    for (long long int i = 0; i < n; i++) {
      long long int mx = a[i] + 5;
      auto idx = upper_bound(a.begin(), a.end(), mx);
      if (idx != a.end()) {
        long long int lst = idx - a.begin();
        long long int gap = lst - i;
        ans = max(ans, gap);
      } else {
        long long int gap = n - i;
        ans = max(ans, gap);
      }
    }
    cout << ans << endl;
  }
  return 0;
}