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)

# /A のために A - 1回

class PriorityQueue
  def initialize(array = [])
    @data = []
    array.each{|a| add(a)}
  end

  def add(element)
    @data.push(element)
    bottom_up
  end

  def peek
    @data[0]
  end

  def poll
    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

  def size
    @data.size
  end

  private

  def swap(i, j)
    @data[i], @data[j] = @data[j], @data[i]
  end

  def bottom_up
    idx = size - 1
    while(has_parent?(idx))
      parent_idx = parent_idx(idx)
      if @data[parent_idx][:key] > @data[idx][:key]
        swap(parent_idx, idx)
        idx = parent_idx
      else
        return
      end
    end
  end

  def top_down
    idx = 0
    while (has_child?(idx))
      a = left_child_idx(idx)
      b = right_child_idx(idx)
      if @data[b].nil?
        c = a
      else
        c = @data[a][:key] <= @data[b][:key] ? a : b
      end
      if @data[idx][:key] > @data[c][:key]
        swap(idx, c)
        idx = c
      else
        return
      end
    end
  end

  def parent_idx(idx)
    (idx - (idx.even? ? 2 : 1)) / 2
  end

  def left_child_idx(idx)
    (idx * 2) + 1
  end

  def right_child_idx(idx)
    (idx * 2) + 2
  end

  def has_child?(idx)
    ((idx * 2) + 1) < @data.size
  end

  def has_parent?(idx)
    idx > 0
  end
end

N, K = gets.split.map(&:to_i)
A = gets.split.map(&:to_i)

queue = PriorityQueue.new()

A.each do |v|
  queue.add({key: -v, value: [v, 1]})
end

K.times do
  v = queue.poll()
  k = (v[:value][0] / ((v[:value][1] + 1) + 0.0)).ceil
  queue.add({key: -k, value: [v[:value][0], v[:value][1] + 1]})
end

puts -queue.poll[:key]