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;
typedef long long ll;

int main(){
 ll n,k;
 cin>>n>>k;
 vector<ll> v(n);
 ll end=INT_MIN;
 
 for(ll i=0;i<n;i++){
 	cin>>v[i];
 	end=max(end,v[i]);
 }
 
 ll start=0;
 ll ans=end;
 while(start<=end){
 	ll mid=(start+end)/2;
 	ll count=0;
 	for(ll i=0;i<n;i++){
 		count+=v[i]/mid;
		if(v[i]%mid==0){
 			count--;
		}	
	}
	if(count>k){
		start=mid+1;
	}
	else{
		ans=mid;
		end=mid-1;
	}
 }
 cout<<ans<<endl;
 return 0;
}