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: Time Limit Exceeded)

use std::{cmp::Ordering, collections::binary_heap};

macro_rules ! input_inner {($ next : expr ) => {} ; ($ next : expr , ) => {} ; ($ next : expr , $ var : ident : $ t : tt $ ($ r : tt ) * ) => {let $ var = read_value ! ($ next , $ t ) ; input_inner ! {$ next $ ($ r ) * } } ; }
macro_rules ! read_value {($ next : expr , ($ ($ t : tt ) ,* ) ) => {($ (read_value ! ($ next , $ t ) ) ,* ) } ; ($ next : expr , [$ t : tt ; $ len : expr ] ) => {(0 ..$ len ) . map (| _ | read_value ! ($ next , $ t ) ) . collect ::< Vec < _ >> () } ; ($ next : expr , chars ) => {read_value ! ($ next , String ) . chars () . collect ::< Vec < char >> () } ; ($ next : expr , usize1 ) => {read_value ! ($ next , usize ) - 1 } ; ($ next : expr , $ t : ty ) => {$ next () . parse ::<$ t > () . expect ("Parse error" ) } ; }
macro_rules ! input {(source = $ s : expr , $ ($ r : tt ) * ) => {let mut iter = $ s . split_whitespace () ; let mut next = || {iter . next () . unwrap () } ; input_inner ! {next , $ ($ r ) * } } ; ($ ($ r : tt ) * ) => {let stdin = std :: io :: stdin () ; let mut bytes = std :: io :: Read :: bytes (std :: io :: BufReader :: new (stdin . lock () ) ) ; let mut next = move || -> String {bytes . by_ref () . map (| r | r . unwrap () as char ) . skip_while (| c | c . is_whitespace () ) . take_while (| c |! c . is_whitespace () ) . collect () } ; input_inner ! {next , $ ($ r ) * } } ; }

#[derive(Debug, PartialEq, Eq)]
struct Cut {
    a: u32,
    k: u32,
}

impl Cut {
    fn devide(&self) -> u32 {
        (self.a + self.k - 1) / self.k
    }
}

impl Ord for Cut {
    fn cmp(&self, other: &Self) -> Ordering {
        (self.devide()).cmp(&(other.devide()))
    }
}
impl PartialOrd for Cut {
    fn partial_cmp(&self, other: &Cut) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

fn main() {
    input!(n: usize, k: usize, a: [u32; n]);
    let mut cut_vec = binary_heap::BinaryHeap::new();
    for ai in a {
        cut_vec.push(Cut { a: ai, k: 1 });
    }

    for _ in 0..k {
        let mut c = cut_vec.pop().unwrap();
        c.k += 1;
        cut_vec.push(c);
    }

    println!("{}", cut_vec.pop().unwrap().devide());
}