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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long max3(long long x, long long y, long long z) {
  return max(max(x, y), z);
}
long long min3(long long x, long long y, long long z) {
  return min(min(x, y), z);
}
const long long N = 1e5 + 10, M = 2e5 + 10, M2 = 1e6 + 10, mod = 1e9 + 7,
                inf = 1e17 + 10;
const long long INF = 1e9 + 7;
void add(long long& a, long long b) {
  a += b;
  if (a >= mod) {
    a -= mod;
  }
}
long long X[] = {0, 1, 0, -1};
long long Y[] = {-1, 0, 1, 0};
long long power(long long x, long long n) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % mod;
    x = ((x % mod) * (x % mod)) % mod;
    n = n / 2;
  }
  return result;
}
const long long Z = 5010;
long long n, k;
long long a[Z];
long long pre[Z];
long long cache[Z][Z];
long long dp(long long idx, long long take) {
  if (take < 0) return -inf;
  if (idx == 0) return 0;
  long long& ans = cache[idx][take];
  if (ans != -1) return ans;
  ans = max(dp(idx - 1, take), dp(pre[idx], take - 1) + idx - pre[idx]);
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 1; i < n + 1; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  long long l = 1;
  for (long long r = 1; r <= n; r++) {
    while (a[r] - a[l] > 5) l++;
    pre[r] = l - 1;
  }
  memset(cache, -1, sizeof(cache));
  long long res = dp(n, k);
  cout << res;
  return 0;
}