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)

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

int main() {
  long N,K;cin>>N>>K;
  vector<long> A(N);
  for(int i=0;i<N;i++) cin>>A.at(i);
  sort(A.begin(),A.end());
  long count;
  long left=0;
  long right=A.at(N-1);
 // int ans;
  while(left<right){
    count=0;
    long X=(left+right)/2;
    for(int i=0;i<N;i++){
      count += ceil(A.at(i)/X);
      if(ceil(A.at(i)/X)*X==A.at(i)) count--;
    }
    //for(int i=0;i<N;i++) count += (A.at(i)+X-1)/X;
    if(count>K) left=X+1;
    else right=X;
  }
  cout<<left<<endl;
  //cout<<left<<endl;
  //for(int i=0;i<N;i++) cout<<ceil(A.at(i)/left)<<endl;
}