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 PHP (7.4.4) (judge verdict: Time Limit Exceeded)

<?php
list($n, $k) = ints();
$a = ints();
// 全ての丸太をx未満の長さにするのに必要な回数がk以下となる時のxの最小値を求める
sort($a);
$ok = $a[$n - 1];
$ng = 0;
while (abs($ok - $ng) > 1) {
    $mid = intdiv($ok + $ng, 2);
    $_mid = $mid;
    $cnt = $_cnt = 0;
    for ($i = 0; $i < $n; ++$i) {
        while ($_mid * $_cnt < $a[$i]) ++$_cnt;
        $cnt += $_cnt - 1;
    }
    if ($cnt <= $k) $ok = $mid;
    else $ng = $mid;
}
echo $ok;
function ints()
{
    return array_map('intval', explode(' ', trim(fgets(STDIN))));
}