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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int main(int argc, const char *argv[]) {
  ll n, k;
  cin >> n >> k;
  vector<ll> va(n);
  for (int i = 0; i < n; ++i) {
    cin >> va[i];
  }

  ll left = 0, right = *max_element(va.begin(), va.end());
  auto check = [](vector<ll> &va, ll mid, ll k) {
    ll cnt = 0;
    for (int i = 0; i < va.size(); ++i) {
      ll a = va[i];
      cnt += a / mid - (a % mid == 0);
    }

    return cnt <= k;
  };

  while (left < right) {
    ll mid = (left + right) / 2;
    if (check(va, mid, k)) {
      right = mid;
    } else {
      left = mid + 1;
    }
  }

  cout << right << '\n';

  return 0;
}