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: Accepted)

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<bitset>
#include<vector>
#include<queue>
//#include<ext/pb_ds/priority_queue.hpp>
#include<set>
#include<map>
#include<math.h>
#include<set>
#define maxn 200005
#define lson (now<<1)
#define rson ((now<<1)|1)
//#define mid ((nl+nr)>>1)
typedef long long ll;
using namespace std;
//using namespace __gnu_pbds; 
const double eps=1e-5;
int n,k;
double a[maxn];
int check(double now)
{
	int cnt=0;
	for(int i=1;i<=n;i++)
	{
		cnt+=(int)(a[i]/now)-1;
		if(a[i]-((int)(a[i]/now)*now)>eps)
		cnt++;
	}
	return cnt<=k;
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
	scanf("%lf",&a[i]);
	double nl=0+eps,nr=1e9;
	while(nr-nl>eps)
	{
		double mid=((nl+nr)/2);
		if(check(mid))
			nr=mid;
		else
			nl=mid;
	}
	printf("%.0lf\n",nl+0.5);
}