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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
#define idfc ios_base::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
//:/
#define pb emplace_back
#define mp make_pair
#define nt _ll128
#define ld long double
long double PI =3.14159265358979323846;
using ll = long long;



const ll modo=1e9+7;
const ll inf=1e17;
const ll ms=(5e5)+5;

int ar[ms];
int n,k;

int peil(int a,int b)
{
    if(a%b)return a/b+1;
    return a/b;
}

int check(int tt)
{
    ll p=0;

    for(int i=1;i<=n;i++)
    {
        p+=peil(ar[i],tt);
        p--;
    }
    if(p<=k)return 1;
    return 0;
}

int main()
{
    idfc;

    cin>>n>>k;
    int i;
    for(i=1;i<=n;i++)cin>>ar[i];
    int l=1;
    int r=1e9;
    int mid;
    int ans=r;
    while(l<=r)
    {
        mid=(l+r)>>1;

        if(check(mid))
        {
            ans=min(ans,mid);
            r=mid-1;
        }
        else l=mid+1;
    }
    cout<<ans;


    return 0;
}