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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
using ll = long long;

template <class T, class F>
T LowerBound(T i0, T i1, F f) {
    while (i0 < i1) {
        T i = i0 + (i1 - i0) / 2;
        if (f(i)) i1 = i; else i0 = i + 1;
    }
    return i0;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, k;
    cin >> n >> k;

    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }

    auto d = LowerBound<int>(1, (int)1e+9 - 1, [&](auto d) {
        //長さd以下を達成できるか
        int s = 0;
        for (int i = 0; i < n; i++) {
            int t = (a[i] + d - 1) / d - 1;
            s += t;
        }
        return s <= k;
    });

    cout << d << endl;

    return 0;
}