Problem:
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You are given $n$ packages of $w_i$ kg from a belt conveyor in order ($i = 0, 1, ... n-1$). You should load all packages onto $k$ trucks which have the common maximum load $P$. Each truck can load consecutive packages (more than or equals to zero) from the belt conveyor unless the total weights of the packages in the sequence does not exceed the maximum load $P$.

Write a program which reads $n$, $k$ and $w_i$, and reports the minimum value of the maximum load $P$ to load all packages from the belt conveyor.

Input

In the first line, two integers $n$ and $k$ are given separated by a space character. In the following $n$ lines, $w_i$ are given respectively.

Output

Print the minimum value of $P$ in a line.

Constraints

$1 \leq n \leq 100,000$

$1 \leq k \leq 100,000$

$1 \leq w_i \leq 10,000$

Sample Input 1

5 3
8
1
7
3
9

Sample Output 1

10

If the first truck loads two packages of $\{8, 1\}$, the second truck loads two packages of $\{7, 3\}$ and the third truck loads a package of $\{9\}$, then the minimum value of the maximum load $P$ shall be 10.

Sample Input 2

4 2
1
2
2
6

Sample Output 2

6

If the first truck loads three packages of $\{1, 2, 2\}$ and the second truck loads a package of $\{6\}$, then the minimum value of the maximum load $P$ shall be 6.

Solution in Ruby (judge verdict: Memory Limit Exceeded)


def count_necessary_num_of_track(baggages, capacity)
  num_of_track = 1
  baggage_index = 0
  loaded_weight = 0
  num_of_baggage = baggages.count
  while baggage_index < num_of_baggage
    loaded_weight += baggages[baggage_index]
    if loaded_weight > capacity
      num_of_track += 1
      loaded_weight = 0
      next
    else
      baggage_index += 1
    end
  end
  return num_of_track
end


def calc_minimum_capacity(baggages, k)
  max_weight = baggages.max
  candidates_for_capacity = (max_weight..10000*10000).to_a #?????§????????????*?????§???????????°=?????§????????????
  return candidates_for_capacity.bsearch{|candidate_for_capacity|
    count_necessary_num_of_track(baggages, candidate_for_capacity) <= k}
end

n, k = gets.chomp.split.map(&:to_i)
baggages = Array.new
n.times do
  baggages.push(gets.chomp.to_i)
end
puts calc_minimum_capacity(baggages, k)