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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

#include <iostream>
#include <fstream>

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define mod1 1000000007
#define mod2 1000000009
int getTheAnswer() {
    return 42;
}

using namespace std;
using ll=long long;

class ELogs {
public:
	void solve(std::istream& cin, std::ostream& cout) {
        ll n,k;
        cin>>n>>k;
        ll a[n];
        for (int i=0;i<n;i++)
            cin>>a[i];
        ll mid,hi=2000000000,lo=1,ans=*max_element(a,a+n);
        while (hi>=lo){
            mid=(hi+lo)>>1;
            ll cuts=0;
            for (int i=0;i<n;i++){
                cuts+=a[i]/mid;
            }
//            cout<<mid<<" "<<cuts<<endl;
            if (cuts<=k){
                hi=mid-1;
                ans=min(mid,ans);
            }
            else lo=mid+1;
        }
        cout<<ans;
	}
};


int main() {
	ELogs solver;
	std::ios::sync_with_stdio(false);
	std::cin.tie(NULL);
	std::cout.tie(NULL);
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
	return 0;
}