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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>

using std::cin;using std::cout;using std::cerr;using std::endl;

#define rep(i,a,b) for(int i = a;i <= b;++i)
#define per(i,a,b) for(int i = a;i >= b;--i)

const int mmax = 2e5;
int n,k;
int a[mmax + 10],l = 1e9,r = 0,ans;

bool check(int x){
	int num = 0,t;
	rep(i,1,n){
		t = a[i] / x;
		if(t * x == a[i]) num += t - 1;
		else num += t;
	}
	return (num <= k?1:0);
}

int main(){

	std::ios::sync_with_stdio(0);
	cin.tie(0);

	cin>>n>>k;

	rep(i,1,n){
		cin>>a[i];
		r = std::max(r,a[i]);
	}
	l = 1;
	ans = r;

	while(r - l >=1){
		int m = (l + r) / 2;
		if(check(m)){
			ans = m;
			r = m - 1;
		}
		else{
			l = m + 1;
		}
	}

	cout<<ans;

	return 0;
}