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: Wrong Answer)

#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use text_io::*;

const U_INF: usize = 1 << 60;
const I_INF: isize = 1 << 60;

fn main() {
    let n: usize = read!();
    let k: usize = read!();

    let mut a = Vec::with_capacity(n);
    let mut max = 0;
    for _ in 0..n {
        let ai: usize = read!();
        max = (max).max(ai);
        a.push(ai);
    }

    if k == 0 {
        println!("{}", max);
        return;
    }

    let mut left = 1;
    let mut right = 1000000000;

    while right - left > 1 {
        let mid = (right + left) / 2;

        let ki = a.iter().map(|ai| *ai / mid).sum::<usize>();

        if ki > k {
            left = mid;
        } else {
            right = mid;
        }
    }

    println!("{}", right);
}