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)

#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<map>

using namespace std;

long long Xd,Xu,D,N,Y,ans,i,j,K,temp,L,L2,A[200001],X;

int main()
{
  
  cin>>N>>K;
  Xu=1;
  for(i=0;i<N;i++)
  {
    cin>>A[i];
    Xu=max(Xu,A[i]);
  }
  if(K==0)
  {
    cout<<Xu;
    return 0;
  }
  Xd=1;
  X=(Xu+Xd)/2;
  while(Xd<X)
  {
    temp=0;
    for(i=0;i<N;i++)
    {
      temp+=(A[i]+X-1)/X-1;
    }
    
    if(temp<=K)
    {
      Xu=X;
    }
    else
    {
      Xd=X;
    }
    X=(Xu+Xd)/2;
  }
  
  cout<<Xu;
  
  
  
}