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<string>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<queue>
#include<deque>
#include<regex>
#include<stack>
#include<stdio.h>
#include<vector>
#include<set>
#include<map>
#include<iomanip>
#define rep(i,n) for(int i=0;i<n;i++)

typedef int long long ll;
using namespace std;
typedef pair<int,int> P;
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
const ll MOD=1e9+7;
static const int MAX = 100;
static const int INF = (1<<23);
template<class T> T gcd(T a, T b){return b? gcd(b,a%b) : a;}
template<class T> T lcm(T a,T b){return a / gcd(a,b)*b;}

int main(){
    cout<<setprecision(16);
    ll n,k;
    cin>>n>>k;
    vector<double> v(n,0);
    rep(i,n){
        cin>>v[i];
    }
    ll l = 0;
    ll r = 1e9;

    rep(ii,5000){
        ll mid = (l+r)/2.0;
        ll sum = 0;
        bool f= true;

        rep(i,n){
            ll c = v[i];
            if(mid!=0) {
                sum += (c / mid);

                if (c % mid == 0)sum--;

            }else{
                f=false;
            }
        }

        if(sum>k&&f){
            l=mid+1;
        }else{
            r=mid;
        }

    }



    cout<<r<<endl;






    return 0;
}