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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define maxn 200005

ll fastpower (ll b , ll p){
    ll k= p%2 ? b : 1 ;
    return p==0 ? 1 : k*(fastpower(b*b,p/2));
}

ll n,k;
ll a[maxn];

int check(ll len){
	ll res=0;
	for(int i=0;i<n;i++)
		res+=(a[i]-1)/len;
	return res<=k;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	ll t=1;
	//cin>>t;
	while(t--){
		cin>>n>>k;
		ll l=0,r=1e10;
		for(int i=0;i<n;i++){
			cin>>a[i];
			//r=max(r,a[i]);
		}
		ll ans=1e18;
		while(l<=r){
			ll mid=(l+r)/2;
			if(check(mid)){
				r=mid-1;
				ans=min(ans,mid);
			}
			else l=mid+1;
		}
		cout<<ans;
	}
}