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 Java (OpenJDK 11.0.6) (judge verdict: Runtime Error)

import java.util.*;
public class Main{
	public static void main(String[] args){
        Scanner s=new Scanner(System.in);
        int n=s.nextInt();
        long k=s.nextInt();
        long[] arr=new long[n];
        
        long i=1;
        long j=0;

        for(int l=0;l<n;l++){
            arr[l]=s.nextInt();
            j=Math.max(j,arr[l]);
        }

        if(k==0){
            System.out.println(j);
            return;
        }

        long ans=j;
        while(i<=j){
            long mid=(i+j)/2;
            if(isPossible(arr,mid,k)){
                ans=mid;
                j=mid-1;
            }else{
                i=mid+1;
            }
        }

        if(isPossible(arr,ans-1,k)){
            ans=ans-1;
            }
        System.out.println(ans);



    }

    public static boolean isPossible(long[] arr,long res,long k){
        
        long count=0;
        for(int i=0;i<arr.length;i++){
            if(arr[i]<=res){

            }else{
                long nc=arr[i]/res;
                if(arr[i]%res==0){
                    nc--;
                }
                System.out.println(nc);
                count=count+nc;
                if(count>k){
                    return false;
                }
            }
        }
        return true;
    }
}