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)

using namespace std;
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
#include <map>
#include <assert.h>
#include <functional>

typedef long long ll;



int main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    int n, k;
    cin >> n >> k;
    vector<int> v;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        v.push_back(x);
    }
    int lo = 1;
    int hi = 1000000000;
    while (hi - lo > 1) {
        int md = (hi + lo) / 2;
        int tot = 0;
        for (int l : v) {
            tot += (l-1) / md;
        }
        if (tot <= k) hi = md;
        else lo = md;
    }
    cout << hi << endl;
}