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: Wrong Answer)

#include <algorithm>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

using namespace std;
using ll = long long;

int main() {
    int N, K;
    cin >> N >> K;

    vector<double> A(N);
    int m = 0;
    for (int i = 0; i < N; i++) {
        cin >> A[i];
        m = max(m, (int)A[i]);
    }
    if (K == 0) {
        cout << m << endl;
        return 0;
    }

    int left = 1;
    int right = m;
    int output = (1 + m) / 2;
    while (1) {
        ll tmp = 0;
        for (int i = 0; i < N; i++) {
            tmp += ceil(A[i] / (double)output);
        }
        if (tmp - N <= K) {
            right = output;
        } else {
            left = output;
        }
        if (right - left == 1) {
            break;
        }
        output = (left + right) / 2;
    }

    cout << output << endl;

    return 0;
}