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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import kotlin.math.abs
import kotlin.math.ceil

private fun getInts(): IntArray {
    return readLine()!!.split(" ").map {it.toInt()}.toIntArray()
}

fun main() {
    val nk = getInts()
    val n = nk[0]
    val k = nk[1]
    val an = getInts()

    if (k==0){
        print(an.max())
        return
    }

    // バイナリサーチする。
    var left = 0.0
    var right = 1000000000.0
    while(true) {
        if (abs(right -left)<0.000001) {
            print(ceil(right).toInt())
            return
        }
        val mid = (left + right) / 2.0
        if (k >= totCut(an,mid)) {
            // 中央も許容できる。
            right = mid
        } else {
            // 許容できない。
            left = mid
        }
    }
}

// カット回数
fun cuttingNum(loglength:Int,limitLength : Double) :Int {
    return ceil(loglength.toDouble()/ limitLength).toInt() - 1
}

fun totCut(arr :IntArray,input: Double): Int {
    return arr.map {
        cuttingNum(it,input)
    }.sum()
}