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)

use proconio::input;
use std::cmp;
fn main() {
    input!{
        n:usize,k:i64,a:[i64;n],
    }
    let ans=solve(n,k,&a);
    println!("{}",ans);
}

fn ceil(a:i64,b:i64)->i64{
    if b==0{
        return 0;
    }
    (a+b-1)/b
}
fn solve(n:usize,k:i64,a:&[i64])->i64{
    let mut max_a:i64=0;
    for i in 0..n{
        max_a=cmp::max(max_a,a[i]);
    }
    let mut right:i64=max_a;
    let mut left:i64=0;
    while right-left>1{
        let x:i64=(right+left)/2;
        let mut cut_times:i64=0;
        for i in 0..n{
            cut_times+=ceil(a[i],x)-1;
        }
        if cut_times>k{
            left=x;
        }
        else {
            right=x;
        }
    }
    right
}