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;
use std::collections::BinaryHeap;

#[derive(Eq, PartialEq)]
struct LOG(u64, u64);

impl Ord for LOG {
    fn cmp(&self, other: &LOG) -> Ordering {
        (self.0 * other.1).cmp(&(self.1 * other.0))
    }
}

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

fn main() {
    let (n, k): (usize, usize) = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let mut iter = buf.split_whitespace();
        (
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
        )
    };

    let a: Vec<u64> = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        let iter = buf.split_whitespace();
        iter.map(|x| x.parse().unwrap()).collect()
    };

    let mut heap = BinaryHeap::new();
    for i in 0..n {
        heap.push(LOG(a[i], 1));
    }

    for _ in 0..k {
        let LOG(x, c) = heap.pop().unwrap();
        heap.push(LOG(x, c + 1));
    }

    let LOG(x, c) = heap.pop().unwrap();
    let ans = (x + c - 1) / c;
    println!("{}", ans);
}