Problem:
Score : 500 points

Problem Statement

We have N logs of lengths A_1,A_2,\cdots A_N.

We can cut these logs at most K times in total. When a log of length L is cut at a point whose distance from an end of the log is t (0<t<L), it becomes two logs of lengths t and L-t.

Find the shortest possible length of the longest log after at most K cuts, and print it after rounding up to an integer.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 10^9

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_N

Output

Print an integer representing the answer.

Sample Input 1

2 3
7 9

Sample Output 1

4

First, we will cut the log of length 7 at a point whose distance from an end of the log is 3.5, resulting in two logs of length 3.5 each.

Next, we will cut the log of length 9 at a point whose distance from an end of the log is 3, resulting in two logs of length 3 and 6.

Lastly, we will cut the log of length 6 at a point whose distance from an end of the log is 3.3, resulting in two logs of length 3.3 and 2.7.

In this case, the longest length of a log will be 3.5, which is the shortest possible result. After rounding up to an integer, the output should be 4.

Sample Input 2

3 0
3 4 5

Sample Output 2

5

Sample Input 3

10 10
158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202

Sample Output 3

292638192

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define endl "\n"

using namespace std;
using ll = long long;

const ll MOD = 1000000007LL; // = 10^9 + 7
const double PI = 3.14159265358979;

void solve()
{
  ll n, k;
  cin >> n >> k;
  vector<ll> a(n);
  for(int i = 0; i < n; ++i)
  {
    cin >> a[i];
  }

  if (k == 0)
  {
    cout << *max_element(a.begin(), a.end());
    return;
  }

  auto f = [&](ll arg)
  {
    ll cnt = 0;
    for(int i = 0; i < n; ++i)
    {
      cnt += a[i] / arg;
      if (a[i] % arg == 0) cnt--;
    }
    return cnt <= k; 
  };

  ll le = 1, re = 1000000001LL;
  while(re - le > 1)
  {
    ll mid = (le + re) / 2;
    if ( f(mid) ) re = mid;
    else le = mid;
  }

  cout << re;
}

int main()
{
  fastio;
  solve();

  return 0;
}