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)

#ifdef DEBUG
#include "../../lib/utils.cpp"
#endif
#include <bits/stdc++.h>

using namespace std;
using ll = int64_t;
#define int int64_t
#define INF 10e9

/* ==== declarations ====== */
// functions
void solve();
void getInput();

// global variables
int N, K;
int maxsize = 0;
vector<int> A;

void getInput()
{
    cin >> N >> K;
    for (int i = 0; i < N; ++i)
    {
        int inp;
        cin >> inp;
        A.push_back(inp);
        maxsize = max(maxsize, inp);
    }
}

bool isOK(double u)
{
    int cost = 0;
    for (auto a : A)
    {
        if (a <= u)
        {
            continue;
        }

        double cc = a / u;
        int c = cc;
        // print("a: ", a, "u: ", u);
        // print("cc:", cc, "c:", c);
        cost += c;
        // print(c, cost);
    }
    // print("cut by ", u, ", cost: ", cost);
    return cost <= K;
}

void solve()
{
    // print(A);
    // isOK(3);
    // print(isOK(3));
    // exit(0);

    double left = -1;
    double right = maxsize;
    while (right - left > 0.01)
    {
        // print("===");
        // print(right, left);
        // print("===");
        double mid = left + (right - left) / 2;
        if (isOK(mid))
        {
            right = mid;
        }
        else
        {
            left = mid;
        }
    }

    int ans = ceil(right);
    cout << ans << endl;
}

signed main()
{
    getInput();
    solve();

    return 0;
}