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)

macro_rules! scan_sub {
    ( $itr:ident, $t:ty ) => {
        let mut buf = String::new();
        let $itr = {
            while buf.trim().len() == 0 {
                buf = String::new();
                std::io::stdin().read_line(&mut buf).unwrap();
            }
            buf.split_whitespace().map(|n| n.parse::<$t>().unwrap())
        };
    };
}

macro_rules! scan {
    ( $x:ident; Vec<$t:ty> ) => {
        scan_sub!(itr, $t);
        let $x = itr.collect::<Vec<_>>();
    };
    ( $( $x:ident ),*; $t:ty ) => {
        scan_sub!(itr, $t);
        let mut itr = itr;
        $(
            let $x = itr.next().unwrap();
        )*
    };
}

fn main() {
    scan!(_n, k; usize);
    scan!(array; Vec<usize>);

    let mut left = 0;
    let mut right = *array.iter().max().unwrap();

    while left < right {
        let m = (left + right) / 2;

        if left == m {
            println!("{}", m+1);
            return;
        }
        if right == m {
            println!("{}", m);
            return;
        }

        let mut s = 0;
        for &a in array.iter() {
            s += if a % m == 0 {
                a / m - 1
            } else {
                a / m
            };
        }

        if s <= k {
            right = m;
        } else {
            left = m;
        }
    }

    let m = (left + right) / 2;
    println!("{}", m);
}