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 Ruby (2.7.1) (judge verdict: Runtime Error)

n,k = gets.chomp.split(" ").map(&:to_i)
a = gets.chomp.split(" ").map(&:to_i)
d = a.max / 2
v = d
if k == 0
  puts a.max
  exit
end
while d != 0
  cnt = 0
  a.each do |x|
    cnt += (x-1) / v
  end
  d = d / 2
  if k >= cnt
    v -= d
  else
    v += d
  end
end
cnt = 0
a.each do |x|
  cnt += (x-1) / v
end
if k < cnt
  v += 1
else
  v -= 1
  a.each do |x|
    cnt += (x-1) / v
  end
  if k < cnt
    v += 1
  end
end
puts v