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)

#![allow(unused_imports)]
use std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use num_traits::clamp;
use ordered_float::OrderedFloat;
use proconio::{input, marker::*, fastout};
use superslice::*;

fn main() {
    input! {
        n: usize, k: usize,
        mut a: [f64; n],
    }

    a.sort_by_key(|&v| Reverse(OrderedFloat(v)));

    if k == 0 {
        println!("{}", a[0].ceil());
        return;
    }

    let mut ng = 0.0f64;
    let mut ok = a[0];
    'out: while ok - ng > 0.000001f64 {
        let mid = (ok + ng) / 2.;
        let mut count = 0;
        for &v in &a {
            count += (v / mid).ceil() as usize - 1;
            if count > k {
                ng = mid;
                continue 'out;
            }
        }
        ok = mid;
    }
    let ans = if (ok - ok.round()).abs() < 0.00001f64 {
        ok.round() as usize
    } else {
        ok.ceil() as usize
    };
    println!("{}", ans);
}