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;

#define ll long long

bool kmax(vector<ll> &v,ll log,ll k) {
	ll sum = 0;
	for(int i = 0;i < v.size()	;i++)
		sum += (v[i]-1)/log;
	return sum<=k;
}
ll bs(vector<ll> &v,ll low,ll high,ll k) {
	if(low < high) {
		ll mid = (high+low)/2;
		if(kmax(v,mid,k))
			return bs(v,low,mid,k);
		else
			return bs(v,mid+1,high,k);
	}
	return low;
}
int main() {
	ll n,k;
	cin >> n >> k;
	vector<ll> v(n);
	ll max_val = 0;
	for(int i = 0;i<n;i++)
	{
		cin >> v[i];
		if(max_val < v[i])
			max_val	= v[i];
	}
	cout << bs(v,0,max_val,k) << endl;
	return 0;
}