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)

#![allow(non_snake_case)]
use std::io::*;
use std::str::FromStr;


fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
    .bytes()
    .map(|c| c.expect("failed to read char") as char)
    .skip_while(|c| c.is_whitespace())
    .take_while(|c| !c.is_whitespace())
    .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let n: usize = read();
    let k: i64 = read();
    let a: Vec<usize> = (0..n).map(|_| read()).collect();
    let mut ans = 0;
    let max = *a.iter().max().unwrap();
    let mut left = 0;
    let mut right = max;
    while left <= right {
        let findLength = (left + right) / 2;
        let mut cut = 0;
        for &i in a.iter() {
            cut += ((i as f64) / (findLength as f64)).ceil() as i64 - 1;
        }
        if cut > k {
            left = findLength + 1;
        } else {
            right = findLength - 1;
            ans = findLength;
        }
    }
    print!("{}", ans);
}