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