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: Accepted)

#include<bits/stdc++.h>
#define watch(x) cout << (#x) << " is " << (x) << endl
#define endl "\n"
typedef long long ll;
using namespace std;
int static fast = [](){
  ios::sync_with_stdio(false);
  cin.tie(0); cout.tie(0); return 0;
}();
// freopen("input.txt", "r", stdin);

int main() {
    ll n, k;
    cin >> n >> k;
    vector<double> logs(n, 0);
    for(auto& val: logs)
        cin >> val;
    double l = 1.0;
    double r = *max_element(logs.begin(), logs.end());
    auto bs = [&] (double chunk_size) {
        ll acc = 0;
        for(auto& val: logs) {
            acc += (ll(ceil(val / chunk_size))-1);
        }
        // cout << "size = " << chunk_size << ", cut " << acc << endl;
        return acc;
    };
    while (r - l > 1e-6) {
        double mid = (l + r) / 2;
        auto use = bs(mid);
        if (use <= k)
            r = mid;
        else
            l = mid;
    }
    cout << ll(ceil(l)) << endl;
    return 0;
}