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 <vector>
#include <string>
#include <cstdlib>
#include<algorithm> 
#include <iomanip>
#include <math.h>
#include <iostream>
#include <bitset>
#include <map>

#define db double
#define vi vector<int>
#define pb push_back
#define fi first
#define se second
#define ll long long
#define ld long double
#define maxn 5005
#define mod 998244353
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vll vector<ll>
#define vpi vector<pii>
#define vpll vector<pll>
#define vv(z,a,b) vector<vector<int>>z(a,vector<int>(b))
#define rep(i,n) for(int i=0;i<(n);i++)
#define all(c) begin(c),end(c)
#define pp pair<int,int>
#define fix(n) fixed << setprecision(n)
#define gmax(a,b) __gcd(a,b)
#define gmin(a,b) a/__gcd(a,b)*b
using namespace std;
//obj;
class binary{
public:
    bool isOK(int index, int key,vi a) {
    if (a[index] >= key) return true;
    else return false;
}
int binary_search(int key,vi a) {
   
    int ng = -1; //「index = 0」が条件を満たすこともあるので、初期値は -1
    int ok = (int)a.size(); // 「index = a.size()-1」が条件を満たさないこともあるので、初期値は a.size()

    /* ok と ng のどちらが大きいかわからないことを考慮 */
    while (abs(ok - ng) > 1) {
        int mid = (ok + ng) / 2;

        if (isOK(mid, key,a)) ok = mid;
        else ng = mid;
    }
    return ok;
}
}obj;

int main(){
int n,k;cin>>n>>k;
vector<double> a(n);
rep(i,n)cin>>a.at(i);
rep(i,k){
sort(all(a));
a.at(n-1)=(a.at(n-1)/2);
}
sort(all(a));
reverse(all(a));
ll ans=a.at(0);
if(a.at(0)>ans)ans++;
cout<<ans<<endl;
return 0;

 }