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: Time Limit Exceeded)

#require 'pp'
require 'set'

class PriorityQueue
    def initialize(array = [])
      @data = []
      array.each{|a| push(a)}
    end
  
    def push(element)
      @data.push(element)
      bottom_up
    end
  
    def pop
      if size == 0
        return nil
      elsif size == 1
        return @data.pop
      else
        min = @data[0]
        @data[0] = @data.pop
        top_down
        return min
      end
    end
  
    private
  
    def swap(i, j)
      @data[i], @data[j] = @data[j], @data[i]
    end
  
    def size
      @data.size
    end
  
    def parent_idx(target_idx)
      (target_idx - (target_idx.even? ? 2 : 1)) / 2
    end
  
    def bottom_up
      target_idx = size - 1
      return if target_idx == 0
      parent_idx = parent_idx(target_idx)
      while (@data[parent_idx][0] > @data[target_idx][0])
        swap(parent_idx, target_idx)
        target_idx = parent_idx
        break if target_idx == 0
        parent_idx = parent_idx(target_idx)
      end
    end
  
    def top_down
      target_idx = 0
  
      # child がある場合
      while (has_child?(target_idx))
  
        a = left_child_idx(target_idx)
        b = right_child_idx(target_idx)
  
        if @data[b].nil?
          c = a
        else
          c = @data[a][0] <= @data[b][0] ? a : b
        end
  
        if @data[target_idx][0] > @data[c][0]
          swap(target_idx, c)
          target_idx = c
        else
          return
        end
      end
    end
  
    # @param Integer
    # @return Integer
    def left_child_idx(idx)
      (idx * 2) + 1
    end
  
    # @param Integer
    # @return Integer
    def right_child_idx(idx)
      (idx * 2) + 2
    end
  
    # @param Integer
    # @return Boolent
    def has_child?(idx)
      ((idx * 2) + 1) < @data.size
    end
end

n, k = gets.split.map(&:to_i)
a_n = gets.split.map(&:to_i)
pq = PriorityQueue.new(a_n.map {|a| [-a.to_f, a, 1] })

#pp pq

k.times do |i|
    #pp pq
    x, orig, cnt = pq.pop
    cnt += 1
    x2 = orig.to_f / cnt
    pq.push([-x2, orig, cnt])

    #pp ordered
end

x = pq.pop
puts -x.first.floor