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>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std ;
#define int                 long long
#define ull                 unsigned long long int
#define ld                  long double
#define vi                  vector<int>
#define vii                 vector<pair<int,int>>
#define vvi                 vector<vector<int>>
#define pii                 pair<int,int>
#define pb                  push_back
#define prec(x)             fixed<<setprecision(15)<<x
#define all(x)              x.begin(),x.end()
#define rall(x)             x.rbegin(),x.rend()
#define rev(x)              reverse(x.begin(),x.end())
#define mset(a,b)           memset(a,b,sizeof(a))
#define mcpy(a,b)           memcpy(a,b,sizeof(a))
#define fo0(i,n)            for(int i=0,i##end=n;i<i##end;i++)
#define fo1(i,n)            for(int i=1,i##end=n;i<=i##end;i++)
#define fo(i,a,b)           for(int i=a,i##end=b;i<=i##end;i++)
#define foe(i,x)            for(__typeof((x).end())i=(x).begin();i!=(x).end();++i)
#define fre(i,x)            for(__typeof((x).rend())i=(x).rbegin();i!=(x).rend();++i)
#define mk(arr,n,type)      type *arr=new type[n];
#define dbg(x) cerr << #x << " is " << x << endl;
#define FAST_IO ios_base::sync_with_stdio(false); cin.tie(0);cout.tie(0)
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
int gcd(int a, int b) { return b ? gcd(b, a%b) : a; }
clock_t beg = clock();
const int inf=1e18;const int mod=1e9+7;const int N=2e5+9;
const double pi = 3.1415926535897932384626433;
void FILE_IO(){
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
}
bool check(double mid,int a[],int n,int k){
int temp=0;
    for(int i=0;i<n;i++){

        if(a[i]>mid){

            temp+=ceil(a[i]*1.0)/mid;

        }
    }
    if(temp>k)
        return false;
    return true;
}
void JH(){
    int d=0,e=0,f=0,g=0,p=0,w=0,flag=0,ans=0,res=0,cnt=0,sum=0;double f1=0.0,f2=0.0,f3=0.0;
    int n,k;cin>>n>>k;
    int a[n];
    fo0(i,n){
        cin>>a[i];
    }
    sort(a,a+n);

    int l=1,h=a[n-1];

    while(l<=h){
        double mid=(l+h)*1.0/2;

        if(check(mid,a,n,k)){
            ans=mid;
            h=ceil(mid)-1;
        }
        else
            l=floor(mid)+1;
    }
    cout<<ans<<endl;

}

int32_t main()
{
    FILE_IO();FAST_IO;int t=1;
    // cin>>t;
    while(t--){JH();}cerr<<fixed<<setprecision(3)<<(double)(clock()-beg)/1000<<"s ";
}