Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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: Wrong Answer)

import scala.io.StdIn.{readInt,readLine}
object Main {
    def main(args: Array[String]): Unit = {
        val input = readLine.split(" ").map(_.toInt) 
        val luggages = (1 to input(0)).foldLeft(Nil: List[Int]) { (list, v) => readInt :: list}
        println(allocation(input(1), luggages.reverse))
    }
    def allocation(k: Int, wList: List[Int]): Int = {
        // 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[Int], truck: Int, capacity: Int, curr: Int): Boolean = luggages match {
            case Nil => true
            case x :: xs => {
                val tmpCurr = curr + x
                if (tmpCurr > capacity) {
                    if (truck == 1) false
                    else inCapacity(xs, truck - 1, capacity, x)
                } else {
                    inCapacity(xs, truck, capacity, curr + x)
                }
            }
        }
        // 最小の最大積載量を2分木探索で探索
        def binSearch(pc: Array[Int], leftPtr: Int, rightPtr: Int, currMin: Int): Int = (leftPtr + rightPtr) / 2 match {
            case center if leftPtr == rightPtr => if (center < currMin && inCapacity(wList, k, pc(center), 0)) pc(center) else currMin
            case center => inCapacity(wList, k, pc(center), 0) match {
                case true => binSearch(pc, leftPtr, center, pc(center))
                case false => binSearch(pc, center + 1, rightPtr, currMin)
            }
        }

        val maxCapa = wList.sum
        val possibleCapa = (1 to maxCapa).toArray
        binSearch(possibleCapa, 0, possibleCapa.length - 1, maxCapa)
    }
}