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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define Mid ((l+r)/2)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define drep(i,a,b) for(int i=(a);i>=(b);--i)
#define file(a) freopen(#a".in","r",stdin),freopen(#a".out","w",stdout);
const int maxn=5e5+5,mod=1e9+7,inf=0x3f3f3f3f;
const double eps=1e-7;
int n,m,Q,K,T;
int read(){
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
double a[maxn];
int dcmp(double x){
    // cout<<x<<endl;
    if(fabs(x)<=eps)return 0;
    return x>0?1:-1;
}
bool pd(double x){
    int sum=0;//cout<<"now"<<x<<endl;
    rep(i,1,n){
        int k=(int)(a[i]/x);//cout<<k<<endl;
        // cout<<k*x<<" "<<a[i]<<endl;
        if(dcmp((double)(k*x-a[i]))<0)++k;
        --k;//cout<<a[i]<<" "<<k<<endl;
        sum+=k;
    }//cout<<x<<" "<<sum<<endl;
    return sum<=K;
}
signed main(){
    //file(a);
    n=read();K=read();double Max=0;
    rep(i,1,n)scanf("%lf",&a[i]),Max=max(Max,a[i]);
    double l=0,r=(double)Max,ans=0;
    while((r-l>=eps)){
        if(pd(Mid))ans=Mid,r=Mid-0.000000001;
        else l=Mid+0.000000001;
    }//cout<<ans<<endl;
    int k=(int)(ans);
    double d=(double)k;
    if(d==ans)printf("%d\n",k);
    else printf("%d\n",k+1);

    return 0;
}