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)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
fn binary_search<T, F>(l: T, r: T, f: F) -> T
where
    T: std::ops::Add<Output = T> + std::ops::Div<Output = T> + PartialEq + From<u8> + Copy,
    F: Fn(T) -> bool,
{
    let mut l = l;
    let mut r = r;

    loop {
        let idx = (l + r) / T::from(2u8);
        // idx == r is required when l > r.
        if idx == l || idx == r {
            break;
        }
        if f(idx) {
            l = idx;
        } else {
            r = idx;
        }
    }

    l
}

fn main() {
    use proconio::input;

    input!(n: usize, k: u64);
    input!(aa: [u64; n]);

    let mut bb = Vec::with_capacity(n);
    let mut max = aa[0];
    for a in aa {
        let a = a * 2;
        bb.push(a);
        if a > max {
            max = a;
        }
    }

    let tt = binary_search(max, 0, |t| {
        // println!("t = {} ---", t);
        let mut cnt = 0;
        for &b in bb.iter() {
            if b % t == 0 {
                cnt += b / t - 1;
            } else {
                cnt += b / t;
            }
            if cnt > k {
                return false;
            }
            // println!("cnt = {}", cnt);
        }
        true
    });

    if tt % 2 == 0 {
        println!("{}", tt / 2);
    } else {
        println!("{}", (tt / 2) + 1);
    }
}
*/

fn main() {
    let exe = "/tmp/bin8488CBF7";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";