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 maxn = 5e3 + 10;
int a[maxn], b[maxn], go[maxn];
struct zkw {
  int N, M;
  int mx[maxn << 2];
  void maintain(int rt) { mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]); }
  void build(int n) {
    for (N = 1; N <= n; N <<= 1)
      ;
    memset(mx, 0, (N + N) * sizeof mx[0]);
    for (int i = 1; i <= N; ++i) mx[i + N] = i;
    for (int i = N - 1; i >= 1; --i) maintain(i);
  }
  void update(int p, int x) {
    mx[p += N] += x;
    for (p >>= 1; p > 0; p >>= 1) maintain(p);
  }
  int point(int p) { return mx[N + p] - p; }
  int query(int l, int r) {
    l--, r++;
    int ret = 0;
    for (l += N, r += N; l ^ r ^ 1; l >>= 1, r >>= 1) {
      if (~l & 1) ret = max(ret, mx[l ^ 1]);
      if (r & 1) ret = max(ret, mx[r ^ 1]);
    }
    return ret;
  }
} dp[2];
int main() {
  double c = clock();
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    int p = upper_bound(a + 1, a + n + 1, a[i] + 5) - a;
    go[i] = p - 1;
  }
  int rt = 0;
  dp[rt].build(n + 1);
  for (int z = 1; z <= k; ++z) {
    rt ^= 1;
    dp[rt].build(n + 1);
    for (int i = n; i >= 1; --i) {
      int x = dp[rt ^ 1].query(i + 1, go[i]) - i;
      x = max(dp[rt ^ 1].point(go[i] + 1) + go[i] - i + 1, x);
      x = max(dp[rt].point(i + 1), x);
      dp[rt].update(i, x);
    }
  }
  cout << dp[rt].point(1) << '\n';
  return 0;
}