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)

END { load __FILE__ unless $stdin.eof? }

n, k = gets.split.map!(&:to_i)
a    = gets.split.map!(&:to_i).sort!

best = a.sum(0.0) / (n + k)
kmax = a.sum { |ai| (ai / best).to_i }

if k >= kmax
  a.map! { |a| r = (ai / best).to_i; k -= r; [a.quo(r + 1), r + 1] }.sort!
else
  a.sort!.map! { |a| [a.quo(1), 1] }
end

k.times do
  x, y = a.pop
  x = x * y / (y + 1)
  y = y + 1
  i = a.bsearch_index { |xi, yi| xi >= x } || n - 1
  a.insert(i, [x, y])
end

puts a.pop[0].ceil