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: Wrong Answer)

//author: clorf
//Think twice,Code once.
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<ctime>
#define INF 1e9
using namespace std;
const int maxn=200010;
const double Pi=acos(-1.0);
template<class T>void read(T &x)
{
	x=0;int f=0;char ch=getchar();
	while(ch<'0'||ch>'9') {f|=(ch=='-');ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	x=f?-x:x;
	return;
}
int n,k,a[maxn],num,sum;
double l,r,mid,ans=1000000000.0;
bool check(double mid)
{
	int sum=0;
	for(int i=1;i<=n;i++)
		sum+=floor(a[i]/mid);
	if(sum<=k)
		return true;
	return false;
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	l=0.0,r=1000000000.0;
	while(r-l>0.001)
	{
		mid=(l+r)/2;
		if(check(mid))
		{
			ans=min(ans,mid);
			r=mid;
		}
		else
			l=mid;
	}
	printf("%d",floor(ans));
	return 0;
}