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 <iostream>
#include<bits/stdc++.h>
#define int long long int
#define vi vector<int>
#define vvi vector<vector<int>>
using namespace std;
int N=1e6+1;
int n,k;
vi a(N);
int m(int l,int h)
{
    return l+(h-l)/2;
}
int bs(int l,int h)
{
    if((h-l)<=1)
    return l+1;
    int mid=m(l,h);int i,count=0;
    for(i=0;i<n;i++)
    {
        if(mid<a[i])
        {
            int cnt=(a[i]-1)/mid;
            count+=(cnt);
            
        }
    }
    if(count>k)
    return bs(mid,h);
    return bs(l,mid);
}
int32_t main() {
	// your code goes here
	int i,j;
cin>>n>>k;
for(i=0;i<n;i++)
cin>>a[i];
sort(a.begin(),a.begin()+n);
cout<<bs(0,a[n-1])<<"\n";
	return 0;
}