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)

fn check(marutas: &Vec<f64>, size: f64, k: usize) -> bool {
    let mut t = 0;
    for &m in marutas {
        if m > size {
            t += (m / size).floor() as usize;
        }
        // m をsize以下にしたい
    }
    // println!("{:?}", (t, size));
    t <= k
}

fn solve() {
    let (n, k): (usize, usize) = (read::<usize>(), read::<usize>());
    let mut a: Vec<f64> = vec![];
    let mut ok: f64 = 0.0;
    for _ in 0..n {
        let v = read();
        a.push(v);
        ok = ok.max(v);
    }
    let mut no = 0.0;
    while ok - no >= 0.0001 {
        // dbg!(no, ok);
        let m = (ok + no) / 2.;
        if check(&a, m, k) {
            ok = m;
        } else {
            no = m;
        }
    }
    println!("{}", ok.ceil());
}

fn main() {
    let stack_size = 104_857_600;
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}

// =========
#[allow(unused_imports)]
use std::cmp::{max, min, Reverse};
#[allow(unused_imports)]
use std::collections::{BinaryHeap, HashMap, HashSet};
#[allow(unused_imports)]
use std::process::exit;

#[allow(dead_code)]
const MOD: usize = 1000000007;

fn read<T: std::str::FromStr>() -> T {
    use std::io::Read;
    let stdin = std::io::stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

// =========