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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

from fractions import Fraction
from math import ceil
from heapq import heappush, heappop

N, K = map(int, input().split())

A_queue = []
for a in input().split():
	a = Fraction(int(a), 1)
	heappush(A_queue, (-a, a, 1))

# search smallest i such that f(i) == True
def auto_bisearch_smallest(f, s = None):
	if s == None:
		s = 0
		w = -1
		while f(s) == True:
			s = w
			w *= 2
	e = s
	w = 1
	while f(e) == False:
		e = s + w
		w *= 2
	
	return bisearch_smallest(f, s, e + 1)

# search smallest i ( s <= i < e ) such that f(i) == True
def bisearch_smallest(f, s, e = None):
	if e == None:
		e = s - 1
		s = 0
	else:
		e -= 1
	
	while s < e:
		m = (s + e) // 2
		if f(m):
			e = m
		else:
			s = m + 1
	
	if s == e and f(s):
		return s
	else:
		return -1

def calc(K):
	while K > 0:
		length, a, div = heappop(A_queue)
		if len(A_queue) > 0:
			ref = A_queue[0][0]
			new_div = auto_bisearch_smallest(lambda div: -a / div > ref, div)
			K -= (new_div - div)
			heappush(A_queue, (-a / new_div, a, new_div))
		else:
			div += K
			K = 0
			heappush(A_queue, (-a / div, a, div))
	return -A_queue[0][0]

print(ceil(calc(K)))