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;
const long long maxn=2e5+5;
const long long inf=0x3f3f3f3f;
long long n,k;
long long a[maxn];
bool check(long long x)
{
	long long sum=0;
	for(long long i=1;i<=n;i++)
	{
		sum+=a[i]/x;
		if(a[i]%x==0)
			sum--;
	}
	if(sum>k)
		return 0;
	else
		return 1;
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	cin>>n>>k;
	for(long long i=1;i<=n;i++)
		cin>>a[i];
	long long l=0,r=1e9;
	while(l<r)
	{
		long long mid=(l+r)/2;
		if(check(mid))
			r=mid;
		else
			l=mid+1;
	}
	cout<<l<<endl;
	return 0;
}