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 proconio::marker::{Chars, Usize1};
use std::iter::Iterator;
use std::collections::*;
use std::cmp::*;
use itertools::Itertools;
use ordered_float::*;

#[allow(unused_macros)]
macro_rules! max {
    ($x:expr) => {
      $x
    };
    ($x:expr, $($xs:tt)+) => {
      max($x,max!($($xs)+))
    };
  }
  
  #[allow(unused_macros)]
  macro_rules! min {
    ($x:expr) => {
      $x
    };
    ($x:expr, $($xs:tt)+) => {
      min($x,min!($($xs)+))
    };
  }

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}



fn main() {
    input! {
    n: usize,
    k: i64,
    a: [i64;n]
    }

    let mut l = 0;
    let mut r = 10000000000;

    while  r - l > 1 {
      let x = (l+r)/2;
      if check (x, &a, n, k){
        r = x;
      }else {
        l = x;
      }
    }

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

fn check (x:i64, a:&Vec<i64>, n:usize, k:i64) -> bool {
  let mut now = 0_i64;
  for i in 0..n {
    now += (a[i] - 1) / x; 
  }
  return now <= k;
}