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: Time Limit Exceeded)

import java.util.*;
public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        // 整数の入力
        int N = sc.nextInt();
        int K = sc.nextInt();
        Double[]A=new Double[N];
        ArrayList<List<Double>> B=new ArrayList<>();
        for(int i=0;i<N;i++){
            A[i]=(double)sc.nextInt();
        }
        Arrays.sort(A, Collections.reverseOrder());
        B.add(Arrays.asList(A));

        ArrayList<Integer> pos=new ArrayList<>();
        pos.add(0);
        ArrayList<Double>now=new ArrayList<>();
        now.add(B.get(0).get(0));
        for(int l=0;l<K;l++){
            double max=-1;
            int maxpos=0;
            for(int i=0;i<now.size();i++){

                if(now.get(i)!=null){
                    if(now.get(i)>max){
                        max=now.get(i);
                        maxpos=i;
                    }
                }
            }
            if(maxpos==B.size()-1){
                B.add(new ArrayList<>());
                pos.add(0);
                now.add(now.get(maxpos)*(maxpos+1)/(maxpos+2));
            }

            B.get(maxpos+1).add(now.get(maxpos)*(maxpos+1)/(maxpos+2));
            pos.set(maxpos,pos.get(maxpos)+1);
            if(pos.get(maxpos)<B.get(maxpos).size()) {
                now.set(maxpos, B.get(maxpos).get(pos.get(maxpos)));
            }else{
                now.set(maxpos, null);
            }

        }
        double max2=0;
        int maxpos2=0;
        for(int i=0;i<B.size();i++){
            if(pos.get(i)<B.get(i).size()) {
                if (B.get(i).get(pos.get(i)) > max2) {
                    max2 = B.get(i).get(pos.get(i));
                    maxpos2 = i;
                }
            }
        }


        System.out.println((int)Math.ceil(max2));
    }
}