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: Time Limit Exceeded)

#include <iostream>
#include <queue>
using namespace std;

int main() {
	int N, K;
	cin >> N >> K;

	auto compare = [](pair<long long, long long>& a, pair<long long, long long>& b) {
		return a.first * b.second < b.first * a.second;
	};

	priority_queue<pair<long long, long long>,
		vector<pair<long long, long long>>,
		decltype(compare)> Q{ compare };
	for (int i = 0; i < N; i++) {
		long long A;
		cin >> A;
		Q.push({ A,1 });
	}
	for (int i = 0; i < K; i++) {
		auto p = Q.top();
		Q.pop();
		p.second++;
		Q.push(p);
	}
	cout << (Q.top().first + Q.top().second - 1) / Q.top().second << endl;
}