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

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n, k;
	cin >> n >> k;
	vector<int> a(n);
	int l = 1, r = 0;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		r = max(r, a[i]);
	}
	while (l < r) {
		int m = (l + r)/2, cuts = 0;
		for (int i = 0; i < n; i++) {
			cuts += (a[i] + m - 1)/m - 1;
		}
		if (cuts <= k) {
			r = m;
		} else {
			l = m + 1;
		}
	}
	cout << r;
}


### 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;

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n, k;
	cin >> n >> k;
	vector<int> a(n);
	int l = 1, r = 0;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		r = max(r, a[i]);
	}
	while (l < r) {
		int m = (l + r)/2, cuts = 0;
		for (int i = 0; i < n; i++) {
			cuts += (a[i] + m - 1)/m - 1;
		}
		if (cuts <= k) {
			r = m;
		} else {
			l = m + 1;
		}
	}
	cout << r;
}


```