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 text_io::*;
use std::process::exit;

fn main() {
    let n:usize = read!();
    let k:usize = read!();
    let mut c:usize = 0;
    let mut e:usize = 0;
    let mut f:usize = 10000000000;
    let mut va:Vec<usize> = Vec::new();
    for i in 0..n {
        let mut a:usize = read!();
        va.push(a);
    }
    if k==0 {
        va.sort();
        println!("{}",va[n-1]);
        exit(0);
    }
    for j in 0..50 {
        c = (e+f)/2;
        let mut d:usize = 0;
        for i in 0..n {
            d += va[i] / c;
        }
        if d <= k {
            f = c;
        } else {
            e = c;
        }
    }
    println!("{}",f);
}