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<stdio.h>
#include<math.h>

int search(int k,int n,int A[]){
	int l=1;
	int r=pow(10,9);
	double x;
	int count;
	int i;

while (r>l){
x=floor((r-l)/2)+l;
    count=0;
	for(i=0;i<n;i++){
	     count+=ceil(A[i]/x)-1;
	}
	if(count<=k){
      r=x;//どちらかは＋１にしておく
	}
	else
	{
		l=x+1;
		if(r==l) x+=1;//最後がNoだと＋１がいる場合分け
	}
}
	
return x;
}

int main(void){
int n,k;
int i;
int ans;
scanf("%d %d",&n,&k);

int A[n];
for(i=0;i<n;i++)
   scanf("%d",&A[i]);



ans=search(k,n,A);
printf("%d\n",ans);



	return 0;
}