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: Accepted)

import java.lang.Long.max
import java.lang.Math.ceil

//

fun main() {
    abc174e()
}

fun abc174e() {
    val (n, k) = readLine()!!.split(" ").map { it.toLong() }
    val a = readLine()!!.split(" ").map { it.toLong() }
    if (k == 0L) {
        println(a.max())
        return
    }
    var total = 0L
    var max = 0L
    for (l in a) {
        max = max(max, l)
        total += l
    }
    if (total <= k) {
        println(1)
        return
    }
    var left = 1L
    var right = max
    while (left + 1L < right) {
        val mid = (left + right) / 2
        var count = 0L
        for (l in a) {
            count += (ceil(l.toDouble() / mid) - 1).toInt().coerceAtLeast(0)
            if (count > k) break
        }
        if (count <= k) {
            right = mid
        } else {
            left = mid
        }
    }
    println(right)
}