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 Python (3.8.2) (judge verdict: Wrong Answer)

from heapq import heappush, heappop, heapify
import math
def solve():
	n , k= map(int, input().strip().split())
	a= [(-int(x),1,int(x)) for x in input().strip().split()]
	heapify(a)
	slices = 0
	while slices < k:
		x,cnt,orig = heappop(a)
		if x == -1:
			return 1
		x = -x-1
		# print(x)
		if (orig+cnt)//(cnt+1) > x:
			if slices +(orig+x-1)//x >= k:
				return x+1
			else:
				cnt_t = (orig+x-1)//x
				slices += cnt_t - cnt
				heappush(a,(-x,cnt,orig))
		else:
			heappush(a,(-((orig+cnt)//(cnt+1)),cnt+1,orig))
			slices+=1
	return -(heappop(a)[0])


print(solve())