#include<bits/stdc++.h>

using namespace std;

int main(){
  int N;
  long long K;
  cin >> N >> K;
  
  vector<int> A(N);
  for(int i = 0; i < N; i++) cin >> A[i];
  
  int ok = 1e9 + 1, ng = 0;
  while(ok - ng > 1){
    int mid = (ok + ng) >> 1;
    
    long long cut = 0;
    for(int i = 0; i < N; i++){
      cut += (A[i] + mid - 1) / mid - 1;
    }
    
    (cut <= K ? ok : ng) = mid;
  }
  
  cout << ok << endl;
  
  return 0;
}
### 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(){
  int N;
  long long K;
  cin >> N >> K;
  
  vector<int> A(N);
  for(int i = 0; i < N; i++) cin >> A[i];
  
  int ok = 1e9 + 1, ng = 0;
  while(ok - ng > 1){
    int mid = (ok + ng) >> 1;
    
    long long cut = 0;
    for(int i = 0; i < N; i++){
      cut += (A[i] + mid - 1) / mid - 1;
    }
    
    (cut <= K ? ok : ng) = mid;
  }
  
  cout << ok << endl;
  
  return 0;
}
```