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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
long double X[200100];
int N;
long long Count(long double len)
{
  long long Ans=0;
  for(int i=0;i<N;i++)
  {
    Ans+=(long long)(X[i]/len);
  }
  return Ans;
}
int main()
{
  long long K;
  cin>>N>>K;
  for(int i=0;i<N;i++)
{
	cin>>X[i];
	X[i]-=0.1;
}
  long double l=1,r=1e9;
  while(l<r-0.1)
  {
    long double mid=(l+r+0.1)/2;
  //  cout<<l<<" "<<r<<" "<<mid<<endl;
    if(Count(mid)<=K)
      r=mid;
    else	l=mid+0.01;
  }
  cout<<(long long)(l+0.999)<<endl;
  return 0;
}