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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <cmath>
using namespace std;

int main(){
    
    long long int i, N, K, min, max, mid, cut;
    cin >> N >> K;
    
    vector<long long int> A(N);
    
    for (i = 0; i < N; i++){
        cin >> A[i];
    }
    
    sort(A.begin(), A.end());
    
    if (K == 0){
        cout << A[N - 1] << "\n";
        return 0;
    }
    
    min = 0;
    max = A[N - 1];
    
    while(max - min > 1){
        
        mid = (min + max) / 2;
        cut = 0;
        
        for (i = 0; i < N; i++){
            cut += (A[i] = 1) / mid;
        }
        
        if (cut > K){
            min = mid;
        } else {
            max = mid;
        }
        
    }
    
    cout << max << "\n";

    return 0;
}