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: Accepted)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::Integer;
#[allow(unused_imports)]
use proconio::derive_readable;
use proconio::fastout;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::cmp;
#[allow(unused_imports)]
use std::collections::*;

#[fastout]
fn main() {
    input! {
        n: usize,
        k: i64,
        a: [i64; n],
    }
    let mut l = 0;
    let mut r = *a.iter().max().unwrap();
    while l + 1 != r {
        let m = (l + r) / 2;
        let cuts: i64 = a.iter().map(|&x| (x + m - 1) / m - 1).sum();
        if cuts <= k {
            r = m;
        } else {
            l = m;
        }
    }

    println!("{}", r);
}