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: Wrong Answer)

import java.util.*;

public class Main {


    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();

        long[] a = new long[n];
        for(int i=0; i<n; i++) {
            a[i] = sc.nextInt();
        }


        long l = 1;
        long r = 1000000000;
        while(r-l>1) {
            long mid = (l+r)/2;
            long num = 0;
            for(int i=0; i<n; i++) {
                long len = mid-1;
                long div = a[i]/len;
                long rem = a[i]%len;
                long add = 0;
                if(rem == 0) {
                    add = Math.max(div-1,0);
                } else {
                    add += Math.max(div, 0);
                }
                num += add;
            }
            if(num > k) {
                l = mid;
            } else {
                r = mid;
            }
        }
        System.out.println(l);

    }
}