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: Accepted)

import java.util.*;

public class  Main 
{
   static  int n , arr[];
   static  long k;
    static boolean check(int x)
    {
        long tot= 0;
        
        for(int i = 0 ; i < n ; i++)
        {
            if(arr[i]%x == 0)
            tot += arr[i]/x - 1;
            
            else
            tot += arr[i]/x;
        }
        
        if(tot <= k)
        return true;
        
        return false;
    }
    
    public static void main(String []args)
    {
        Scanner sc = new Scanner(System.in);
         n = sc.nextInt();
          k = sc.nextLong();
        arr = new int[n];
        for(int i = 0 ;i  < n ; i++)
        {
            arr[i] = sc.nextInt();
        }
        
        int l = 1 , r = 1000000000;
        while(l < r)
        {
            int mid = (l+r)/2;
            if(check(mid))
            r = mid;
            
            else 
            l = mid+1;
        }
        System.out.println(l);
    }
}