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: Accepted)

MAX_N_LOADS = 100000
MAX_WEIGHT  = 10000

def loadable_num(n_trucks, load_weights, capacity)
  w_loaded = 0
  n_trucks_loaded = 0
  load_weights.each_with_index do |weight, i|
    return i if capacity < weight
    w_loaded += weight
    if capacity < w_loaded
      w_loaded = weight
      n_trucks_loaded += 1
      return i if n_trucks == n_trucks_loaded
    end
  end
  load_weights.size
end

def find_min_capacity(n_trucks, load_weights)
  left    = 0
  right   = MAX_N_LOADS * MAX_WEIGHT
  n_loads = load_weights.size
  while left < right
    mid = (left + right) / 2
    n_loadables = loadable_num(n_trucks, load_weights, mid)
    if n_loadables < n_loads
      left = mid + 1
    else
      right = mid
    end
  end
  right
end

n_loads, n_trucks = gets.split.map &:to_i
load_weights = n_loads.times.map do
  gets.chomp.to_i
end

puts find_min_capacity(n_trucks, load_weights)