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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
using namespace std;

#include <math.h>
#include <iomanip>
#include <vector>
#include <queue>
#include <climits>
using Graph = vector<vector<int>>;
const double PI = 3.14159265358979323846;

int needed_kaisu(long long a, long long N, long long *A) {
	int count = 0;
	for (long long i = 0; i < N; i++) {
		count += (A[i] + a - 1) / a;
	}
	return count;
}
int main() {


	long long N,K,A[200100],ans,mem[200100];
	priority_queue<pair<long long, long long>> queue;

	long long max = 0;

	cin >> N >> K;
	for (long long i = 0; i < N; i++) {
		cin >> A[i];
		if (max < A[i]) {
			max = A[i];
		}
	}

	long long lb = 0;
	long long ub = max;

	while (1) {
		long long tmp = (lb + ub) / 2;
		if (needed_kaisu(tmp, N, A) <= K) {
			ub = tmp;
		}
		else {
			lb = tmp;
		}
		if (ub == lb) {
			break;
		}
	}

	cout << ub <<endl;

	cin >> N;

	return 0;


}