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)

#![allow(unused_parens)]
#![allow(unused_imports)]
#![allow(non_upper_case_globals)]
#![allow(non_snake_case)]
#![allow(unused_mut)]
#![allow(unused_variables)]
#![allow(dead_code)]

use itertools::Itertools;
use proconio::input;
use proconio::marker::{Chars, Usize1};

#[allow(unused_macros)]
#[cfg(debug_assertions)]
macro_rules! mydbg {
    //($arg:expr) => (dbg!($arg))
    //($arg:expr) => (println!("{:?}",$arg));
      ($($a:expr),*) => {
          eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
      }
}
#[cfg(not(debug_assertions))]
macro_rules! mydbg {
    ($($arg:expr),*) => {};
}

macro_rules! echo {
    ($($a:expr),*) => {
          $(println!("{}",$a))*
      }
}

use std::cmp::*;
use std::collections::*;
use std::ops::{Add, Div, Mul, Sub};

#[allow(dead_code)]
static INF_I64: i64 = 92233720368547758;
#[allow(dead_code)]
static INF_I32: i32 = 21474836;
#[allow(dead_code)]
static INF_USIZE: usize = 18446744073709551;
#[allow(dead_code)]
static M_O_D: usize = 1000000007;
#[allow(dead_code)]
static PAI: f64 = 3.1415926535897932;

trait IteratorExt: Iterator {
    fn toVec(self) -> Vec<Self::Item>;
}

impl<T: Iterator> IteratorExt for T {
    fn toVec(self) -> Vec<Self::Item> {
        self.collect()
    }
}

trait CharExt {
    fn toNum(&self) -> usize;
    fn toAlphabetIndex(&self) -> usize;
    fn toNumIndex(&self) -> usize;
}
impl CharExt for char {
    fn toNum(&self) -> usize {
        return *self as usize;
    }
    fn toAlphabetIndex(&self) -> usize {
        return self.toNum() - 'a' as usize;
    }
    fn toNumIndex(&self) -> usize {
        return self.toNum() - '0' as usize;
    }
}

trait VectorExt {
    fn joinToString(&self, s: &str) -> String;
}
impl<T: ToString> VectorExt for Vec<T> {
    fn joinToString(&self, s: &str) -> String {
        return self
            .iter()
            .map(|x| x.to_string())
            .collect::<Vec<_>>()
            .join(s);
    }
}

trait StringExt {
    fn get_reverse(&self) -> String;
}
impl StringExt for String {
    fn get_reverse(&self) -> String {
        self.chars().rev().collect::<String>()
    }
}

trait UsizeExt {
    fn pow(&self, n: usize) -> usize;
}
impl UsizeExt for usize {
    fn pow(&self, n: usize) -> usize {
        return ((*self as u64).pow(n as u32)) as usize;
    }
}

fn check(a: &Vec<usize>, m: usize, k: usize) -> bool {
    let mut n = 0;
    for &item in a {
        let c = (item + m - 1) / m;
        n += c - 1;
    }
    mydbg!(n, m);
    if n <= k {
        return true;
    } else {
        return false;
    }
}

fn main() {
    use itertools_num::ItertoolsNum as _;
    input! {
       N: usize,
       K:usize,
       A:[usize;N],
    }

    let mut ans: usize = 0;
    let mut l = 0;
    let mut r = *A.iter().max().unwrap();
    while r - l > 1 {
        let m = (l + r) / 2;
        if check(&A, m, K) {
            r = m;
        } else {
            l = m;
        }
    }

    echo!(r);
}