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 <bits/stdc++.h>

typedef long long ll;

using namespace std;


const ll mod=1000000007;

struct Point{
    ll originallength;
    ll n=0;
    double  length;
    void set(ll k){
        n=k+n;
        length=(double)originallength/(n);
    }
};



int  main(){
    ll N,K;
    cin>>N>>K;
    Point s[N];

    for(int i=0;i!=N;i++){
        cin>>s[i].originallength;
        s[i].set(1);
        cout<<s[i].length<<endl;
        
    }

    sort(s,s+N,[](Point s1,Point s2){return s1.length>s2.length;});

    while(K>0){
        ll p=s[0].length/s[1].length;
        cout<<"p:"<<p<<endl;
        s[0].set(p);
        cout<<s[0].length<<endl;
        sort(s,s+N,[](Point s1,Point s2){return s1.length>s2.length;});
        K=K-p;

    }


    cout<<(ll)ceil(s[0].length)<<endl;

}