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;


long long int Exp(long long int exp_n,long long int exp_e, long long exp_mod);


int main(){
  long long int N,K,Max_A=0,O;
  priority_queue<long long int> q;
  cin >> N >> K;
  long long int A[N];
  for(int i=0;i<N;i++){
    cin >> A[i];
    Max_A = max(Max_A,A[i]);
  }
  long long int X,top=Max_A,btm=0;
  X=(top+btm)/2;
  while(top != X){
    O=0;
    for(int i=0;i<N;i++){
      O=O+(A[i]-1)/X;
    }
    if(O>K){
      btm=X;
      X=(top+btm+1)/2;
    }else{
      top=X;
      X=(top+btm+1)/2;
    }
  }
  
  cout << X <<endl;
  
  return 0;
}


long long int Exp(long long int exp_n,long long int exp_e, long long int exp_mod){
  long long int exp_o = 1;
  long long int exp_x=exp_n;
  long long int exp_tmp;
  
  while(0<exp_e){
    exp_tmp = exp_e/2;
    if(exp_e - exp_tmp*2 == 1){
      exp_o = exp_o*exp_x;
      if(exp_o>exp_mod){
        exp_o=exp_o%exp_mod;
      }
    }
    
    exp_x = exp_x*exp_x;
    if(exp_x>exp_mod){
      exp_x=exp_x%exp_mod;
    }
    exp_e = exp_tmp;
  }
  return exp_o;
}