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>
#define fast ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define endl "\n";
using namespace std;
typedef long long int ll;
bool possible(vector<int>a,double x,int k)
{
   int k0;
	for (int d : a) {
		k0 = ceil(d / (x )) - 1;
		if (round(d / (k0 + 1)) > x) {
			k0++;
		}
		k -= k0;
	}
 
	return k >= 0;
}
int main()
{
    int n,k;
    cin>>n>>k;
    vector<int>a(n);
    for(int i=0;i<n;i++)cin>>a[i];
    int mn =0,mx=(int)1e9+100;
    while(mn!=mx)
    {
        int mid = (mn)+(mx-mn)/2;
        if(possible(a,mid,k))
            mx = mid;
        else
            mn = mid+1;
    }
    cout<<mx<<endl;
    return 0;
}