#include<bits/stdc++.h>
using namespace std;
using ll = long long;

int main()
{
	ll n,k;
	cin >> n >> k;
	vector<ll> a(n);
	for(int i = 0;i<n;i++)cin >> a[i];
	ll l = 0,r = 2*1e9;

	while(r-l>1)
	{
		ll mid = (l+r)/2;
		//最大の長さをmidにできるか
		ll now = 0;
		for(int i = 0;i<n;i++)
		{
			if(a[i]<=mid)continue;
			now += (a[i]+mid-1)/mid;
			now--;
		}
		if(now<=k)r=mid;

		else l = mid;
	}
	cout<<r<<endl;
}

### Prompt

Generate a cpp solution to the following problem:

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.

Examples

Input

2 3
7 9


Output

4


Input

3 0
3 4 5


Output

5


Input

10 10
158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202


Output

292638192

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
using ll = long long;

int main()
{
	ll n,k;
	cin >> n >> k;
	vector<ll> a(n);
	for(int i = 0;i<n;i++)cin >> a[i];
	ll l = 0,r = 2*1e9;

	while(r-l>1)
	{
		ll mid = (l+r)/2;
		//最大の長さをmidにできるか
		ll now = 0;
		for(int i = 0;i<n;i++)
		{
			if(a[i]<=mid)continue;
			now += (a[i]+mid-1)/mid;
			now--;
		}
		if(now<=k)r=mid;

		else l = mid;
	}
	cout<<r<<endl;
}

```