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)

//Krunal_Mathukiya
#include<bits/stdc++.h>
using namespace std;
#define lli long long int
#define li long int
#define pb push_back
#define mkp make_pair
#define nikalL return
#define umap unordered_map
#define fr(i,s,n) for(int i=s;i<=n;++i)
#define frr(i,s,n) for(int i=s;i>=n;--i)
#define jaldiKarNe ios_base::sync_with_stdio(false);cin.tie(NULL);
#define mod 1000000007
int n;
lli arr[200007];
void TestingChaluKarEyy(){
	lli k;
	cin>>n>>k;
	fr(i,1,n) cin>>arr[i];
	lli start=1,end=1000000000;
	lli mid;
	lli ans=10000000000;
	while(start<=end) {
		mid=(start+end)/2;
		// cout<<start<<" "<<end<<" : "<<mid<<"\n";
		lli cnt=0;
		fr(i,1,n) {
			cnt+=(ceil((long double)arr[i]/mid)-1);
		}
		if(cnt>k) start=mid+1;
		else {
			ans=min(ans,mid);
			end=mid-1;
		}
		
	}
	cout<<ans<<"\n";
}
int main(){
	#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif
	jaldiKarNe;
	int TestingKaravvuChhe=1;
	// cin>>TestingKaravvuChhe;
	while(TestingKaravvuChhe--){
		TestingChaluKarEyy();
	}
}