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)

use proconio::input;
//use proconio::marker::{Bytes, Chars};

// x ** n % mods を求める
pub fn mod_pow(x: i64, n: i64, mods: i64) -> i64 {
    if n == 0 {
        return 1;
    }
    let mut res = mod_pow(x * x % mods, n / 2, mods);
    if n % 2 == 1 {
        res = res * x % mods;
    }
    return res;
}

// (1/n) % mods を求める
pub fn mod_inv(n: i64, mods: i64) -> i64 {
    return mod_pow(n, mods-2, mods);
}

// nCk % mods を求める
pub fn mod_comb(n: i64, k: i64, mods: i64) -> i64 {
    let sum1 = (n-k+1..n+1).fold(1, |ac, x| ac * x % mods);
    let sum2 = (1..k+1).fold(1, |ac, x| ac * x % mods);
    return sum1 * mod_inv(sum2, mods) % mods;
}

// nCk の組み合わせを列挙する
pub fn combination(n: usize, k: usize) -> Vec<Vec<usize>> {
    let mut ans = Vec::new();
    if k == 1 {
        for i in 1..n+1 {
            let mut tmp = Vec::new();
            tmp.push(i);
            ans.push(tmp);
        }
    }
    else {
        for iter in combination(n-1, k-1) {
            let max = iter.iter().max().unwrap();
            for j in *max+1..n+1 {
                let mut tmp = iter.clone();
                tmp.push(j);
                ans.push(tmp);
            }
        }
    }
    ans
}


fn main() {
    input!{
        n: usize,
        k: f64,
        a: [f64; n],
    }
    let mut max = a.iter().fold(0.0/0.0, |m, v| v.max(m));
    let mut min = 0.0;
    let mut ans = 0.0;

    if k == 0.0 {
        println!("{}", max);
        return;
    }
    
    while max - min > 1.0 {
        let div = ((max + min) / 2.0).ceil();
        let mut count = 0.0;
        for i in 0..n {
            count += (a[i] / div).ceil() - 1.0;
        }
        if count <= k {
            ans = div;
            max = div;
        }
        else {
            min = div;
        }
    }
    println!("{}", ans);
    return;
}