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: Runtime Error)

use std::collections::*;
use std::io::*;
use std::*;
// template {{{
macro_rules! get {
    ($($t:tt),*; $n:expr) => {
        {
            let stdin = std::io::stdin();
            let ret = std::io::BufRead::lines(stdin.lock()).take($n).map(|line| {
                let line = line.unwrap();
                let mut it = line.split_whitespace();
                _get!(it; $($t),*)
            }).collect::<Vec<_>>();
            ret
        }
    };
    ($($t:tt),*) => {
        {
            let mut line = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut it = line.split_whitespace();
            _get!(it; $($t),*)
        }
    };
}
macro_rules! _get {
    ($it:ident; [char]) => {
        _get!($it; String).chars().collect::<Vec<_>>()
    };
    ($it:ident; [u8]) => {
        _get!($it; String).bytes().collect::<Vec<_>>()
    };
    ($it:ident; usize1) => {
        $it.next().unwrap().parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1
    };
    ($it:ident; [usize1]) => {
        $it.map(|s| s.parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1).collect::<Vec<_>>()
    };
    ($it:ident; [$t:ty]) => {
        $it.map(|s| s.parse::<$t>().unwrap_or_else(|e| panic!("{}", e))).collect::<Vec<_>>()
    };
    ($it:ident; $t:ty) => {
        $it.next().unwrap().parse::<$t>().unwrap_or_else(|e| panic!("{}", e))
    };
    ($it:ident; $($t:tt),+) => {
        ($(_get!($it; $t)),*)
    };
}
// }}}

fn main() {
    let (n, k) = get!(usize, i64);
    let a = get!([i64]);
    let mut l: i64 = -1;
    let mut r: i64 = *a.iter().max().unwrap();
    while r - l > 1 {
        let mid = (l + r) / 2;
        let mut count: i64 = 0;
        for i in 0..n {
            count += a[i] / mid;
        }
        if count <= k {
            r = mid;
        } else {
            l = mid;
        }
    }
    println!("{}", r);
}