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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

using llong = long long int;

#define rep(i, n) for (int i = 0; i < n; ++i)
#define stl_rep(itr, x) for (auto itr = x.begin(); itr != x.end(); ++itr)
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()

const static int MOD = 1000000007;
const static int INF = 1000000000;
const static int dx[4] = {1, 0, -1, 0};
const static int dy[4] = {0, 1, 0, -1};

int main (int argc, char *argv[]) {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, k;
    cin >> n >> k;
    vector<int> logs(n);
    rep(i, n) cin >> logs[i];

    int left = -1, right = INF, mid;
    while (right - left > 1) {
        mid = left + (right - left) / 2;
        int cut = 0;
        rep(i, n) cut += (logs[i] + mid - 1) / mid - 1;
        if (cut <= k) right = mid;
        else left = mid;
    }

    cout << right << endl;

    return 0;
}