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.*;
	import java.util.Map.Entry;
	 
	 
	 class Main {
		 static int mod =  (int) (Math.pow(10,9)+7);
//		 static int mod =  (int) 998244353;
		 static List<ArrayList<Integer>> list= new ArrayList<ArrayList<Integer>>();
		 static int a_dist[];
		 public static void main(String[] args) {
	    	
	        Scanner sc = new Scanner(System.in);
	        int N = sc.nextInt();
	        long K = sc.nextInt();
	        long[] a = new long[N];
	        for (int i=0 ; i<N; i++) {
	            a[i] = Long.parseLong(sc.next());
	        }	        
	        
	        long left=0;
	        long right=mod;
	        while(right-left>1) {
	        	long mid = left + (right-left)/2;
	        	long cnt =0;
	        	for(int i=0;i<N;i++) {	        			
	        		cnt=cnt+a[i]/mid-1;
	        		if(a[i]%mid!=0)cnt++;
	        	}
	        	if(cnt>K) {
	        		left=mid;
	        	} else {
	        		right=mid;
	        	}
	        }
	        
	        
	        
	        System.out.println(right);
		 }
		 
	}