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;
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define lli long long int
#define tc int t;cin>>t;while(t--)
#define all(x) x.begin(),x.end()
#define f first
#define s second
#define pb push_back
#define vi vector<int>
#define vll vector<lli>
#define maxn 1771562
#define mod 1000000007
#define inf 10000000



int main()
{
    fastio;
    int n,k;
    cin>>n>>k;
    lli a[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    lli l=1,h=1e9,ans;
    while(l<=h){
        lli mid=(l+h)/2;
        lli ca=0;
        for(int i=0;i<n;i++){
            ca+=a[i]/mid-1;
            if(a[i]%mid) ca++;
        }
        //cout<<mid<<" "<<ca<<"\n";
        if(ca<=k) ans=mid,h=mid-1;
        else l=mid+1;
    }
    cout<<ans<<"\n";
    return 0;
}