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.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = Integer.parseInt(scan.next());
		int k = Integer.parseInt(scan.next());
		String s = scan.nextLine();
		String[] a = s.split(" ");
		int count = 0;
		int max = 0;

		while(count<k) {
			for(int i=0; i < n; i++) {
				if(max<Integer.parseInt(a[i])) {
					max = i;
				}
			}
			int l = Integer.parseInt(a[max]);
			if(l%2==1) {
				a[max] = Integer.toString(l/2+1);
			}else {
				a[max] = Integer.toString(l/2);
			}
			count++;
		}

		for(int i=0; i < n; i++) {
			if(max<Integer.parseInt(a[i])) {
				max = i;
			}
		}
		System.out.println(a[max]);
		scan.close();
	}

}