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 proconio::{input, fastout};
use proconio::marker::*;
use smallvec::alloc::collections::BinaryHeap;

#[fastout]
fn main() {
    input! {
    n: usize,
    k: usize,
    a: [usize; n]
    }
    let (mut s, mut t) = (0, std::usize::MAX);
    while s + 1 < t {
        let mid = (s + t) / 2;
        if is_ok(&a, k, mid) {
            t = mid
        } else {
            s = mid
        }
    }
    println!("{}", t);
}

fn is_ok(a: &Vec<usize>, k: usize, d: usize) -> bool {
    let mut count = 0;
    for &i in a {
        count += i / d + if i % d == 0 {0} else {1} - 1;
    }
    count <= k
}