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 N = 100005, INF = 2000000000000000000;
long long power(long long a, long long b, long long p) {
  if (a == 0) return 0;
  long long res = 1;
  a %= p;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return res;
}
bool prime[N] = {true};
long long sum[N] = {0};
void pre() {
  memset(prime, true, sizeof(prime));
  for (long long i = 2; i <= N; i++) {
    if (prime[i]) {
      for (long long j = i * i; j <= N; j += i) {
        prime[j] = false;
      }
    }
  }
  for (long long i = 2; i < N; i++) {
    if (prime[i]) sum[i] += i;
    sum[i + 1] = sum[i];
  }
  return;
}
class yup {
 public:
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    return gcd(b % a, a);
  }
  long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
};
long long minCoins(long long coins[], long long m, long long V) {
  if (V == 0) return 0;
  long long res = INT_MAX;
  for (long long i = 0; i < m; i++) {
    if (coins[i] <= V) {
      long long sub_res = minCoins(coins, m, V - coins[i]);
      if (sub_res != INT_MAX && sub_res + 1 < res) res = sub_res + 1;
    }
  }
  return res;
}
bool com(long long a, long long b) { return a > b; }
long long ans = 0;
class gfg {
 public:
  unsigned long long getFirstSetBitPos(long long n) { return log2(n & -n) + 1; }
};
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  long long a[n], ans = -1;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  for (long long i = 0; i < n; i++) {
    long long x = a[i] + 5;
    long long pos = upper_bound(a, a + n, x) - a - 1;
    ans = max(ans, pos - i);
  }
  cout << ans + 1;
}