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 Scala (judge verdict: Accepted)

import scala.io.StdIn.{readLong,readLine}
object Main {
    def main(args: Array[String]): Unit = {
        val input = readLine.split(" ").map(_.toInt) 
        val luggages = (1 to input(0)).foldLeft(Nil: List[Long]) { (list, v) => readLong :: list}
        println(allocation(input(1), luggages.reverse))
    }
    def allocation(k: Int, wList: List[Long]): Long = {
        // 最大積載量 = capacity の場合にトラックに荷物が乗り切るかどうか
        def inCapacity(luggages: List[Long], truck: Int, capacity: Long, curr: Long): Boolean = luggages match {
            case Nil => true
            case x :: xs if (x > capacity) => false
            case x :: xs => curr + x match {
                case tmp if tmp > capacity => if (truck == 1) false else inCapacity(xs, truck - 1, capacity, x)
                case tmp => inCapacity(xs, truck, capacity, tmp)
            }
        }
        // すべての荷物が乗り切るトラックの積載量の最小値を2分木探索で探索
        def binSearch(min: Long, max: Long, currMin: Long): Long = (min + max) / 2 match {
            case median if min == max => if (median < currMin && inCapacity(wList, k, median, 0L)) median else currMin
            case median => inCapacity(wList, k, median, 0) match {
                case true => binSearch(min, median, median)
                case false => binSearch(median + 1, max, currMin)
            }
        }

        val maxCapa = 100000 * 100000
        binSearch(0L, maxCapa, maxCapa)
    }
}