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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {n: usize, k: usize, a:[usize; n]};

    let mut a: Vec<usize> = a;

    let mut cut_count = 0;

    while cut_count < k {
        a.sort();
        let mut base = a[(a.len() + 1) / 2 - 1].clone();

        let mut big_a: Vec<usize> = a.split_off(a.len() / 2);
        // println!("base : {:?}", base);
        // println!("big_a: {:?}", big_a);
        // println!("a    : {:?}", a);

        for big_a_iter in big_a.iter() {
            let big = *big_a.iter().max().unwrap();
            if base * 2 > big {
                a.push(big / 2 + 1);
                a.push(big / 2 + 1);
            } else {
                a.push(base);
                a.push(big_a_iter - base);
            }
            cut_count += 1;
            if cut_count == k {
                break;
            }
        }
    }

    println!("{:?}", a.iter().max());
}