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 <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define FSP(x) fixed << setprecision(x)
using namespace std;
using ll = long long;
constexpr ll INF = LLONG_MAX;
const long double PI = acosl(-1);
void Yes() {cout << "Yes\n";}
void No() {cout << "No\n";}
void YES() {cout << "YES\n";}
void NO() {cout << "NO\n";}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	//constexpr ll P = 1e9 + 7;
	//constexpr ll P = 998244353;

	ll n, k;
	cin >> n >> k;
	vector<ll> a(n);
	for (ll i = 0; i < n; i++) cin >> a[i];
	sort(ALL(a));



	if (k == 0) {
		cout << a[n - 1] << endl;
		return 0;
	}



	if (a[n - 1] == 1) {
		cout << 1 << endl;
		return 0;
	}



	ll l = 1, r = a[n - 1];
	auto is_ok = [&] (ll x) {
		ll lim = k;
		for (auto i : a) {
			lim -= i / x + !!(i % x) - 1;
			if (lim < 0) return false;
		}
		return true;
	};
	while (l + 1 < r) {
		ll mid = (l + r) / 2;
		if (is_ok(mid)) r = mid;
		else l = mid;
	}
	cout << r << endl;
}