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 maxx = 1e6 + 5;
bool dd[maxx], check, d1[maxx];
long long tr[maxx], val[maxx];
vector<long long> adj[maxx];
void bfs(long long u) {
  queue<long long> qu;
  fill_n(dd, maxx, true);
  fill_n(tr, maxx, 0);
  fill_n(val, maxx, 100000);
  dd[u] = false;
  val[u] = 0;
  qu.push(u);
  do {
    long long v = qu.front();
    qu.pop();
    for (long long x : adj[v]) {
      if (dd[x]) {
        tr[x] = v;
        dd[x] = false;
        qu.push(x);
        val[x] = min(val[v] + 1, val[x]);
      }
    }
  } while (!qu.empty());
}
void sol() {
  long long n, c = 0, k, b[maxx], tong = 0, m, x, x1, y1, l, h, t = 0, y;
  cin >> n >> m;
  vector<long long> st[maxx];
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    adj[y].push_back(x);
    st[x].push_back(y);
  }
  cin >> k;
  for (int i = 1; i <= k; i++) {
    cin >> b[i];
  }
  bfs(b[k]);
  for (int i = 1; i <= n; i++)
    if (!dd[i]) cout << val[i] << " " << i << '\n';
  x1 = 0;
  y1 = 0;
  for (int i = 1; i < k; i++) {
    l = 1e9;
    h = 0;
    for (long long j : st[b[i]])
      if (val[j] < l) l = val[j];
    for (long long j : st[b[i]]) {
      if (val[j] == l && j != b[i + 1]) ++h;
    }
    cout << h << " " << b[i] << '\n';
    if (val[b[i + 1]] > l) ++x1;
    y1 += h;
  }
  cout << x1 << " " << y1;
}
long long a[maxx];
long long bs(long long v, long long n) {
  long long l = 1, h = n, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (a[m] > v)
      h = m - 1;
    else
      l = m + 1;
  }
  return h;
}
long long bs1(long long v, long long n) {
  long long l = 0, h = n, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (a[m] < v)
      l = m + 1;
    else
      h = m - 1;
  }
  return l;
}
void cf() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  vector<int> cnt(n);
  for (int i = 0; i < n; ++i) {
    while (i + cnt[i] < n && a[i + cnt[i]] - a[i] <= 5) {
      ++cnt[i];
    }
  }
  vector<vector<int>> dp(n + 1, vector<int>(k + 1));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= k; ++j) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      if (j + 1 <= k) {
        dp[i + cnt[i]][j + 1] = max(dp[i + cnt[i]][j + 1], dp[i][j] + cnt[i]);
      }
    }
  }
  cout << dp[n][k] << endl;
  return;
}
long long uoc(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
bool prime(long long v) {
  if (v < 2) return false;
  for (int i = 2; i <= v / i; i++) {
    if (v % i == 0) return false;
  }
  return true;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cf();
}