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)

//-----------------------
// author : xyqkoala
// time : 2020-08-03
//-----------------------

#include<iostream>
#include<vector>
#include<string>
#include<deque>
#include<unordered_map>
#include<unordered_set>
#include<queue>
#include<algorithm>
#include<cmath>
#include<numeric>
using namespace std;

typedef long long LL;
const int MOD = 1e9+7;

void solve();


int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	int T = 1;
	// cin>>T;
	while(T--){
		solve();
	}
	return 0;
}


LL check(vector<int> &nums,int m){
	LL res = 0;
	for(auto item:nums){
		int cnt = item/m;
		if(item%m) cnt++;
		res += 1ll *(cnt-1);
	}
	return res;
}


void solve(){
	int n,k;
	cin>>n>>k;
	vector<int> nums(n);
	for(int i=0;i<n;i++){
		cin>>nums[i];
	}
	int l=0,r = MOD;
	while(l<r){
		int m = (r-l)/2+l;
		if(check(nums,m) <=1ll*k) r = m;
		else l=m+1;
	}
	cout<<l<<endl;
}