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>
#define ll long long
#define vi vector<int>
#define vll vector<ll>
#define pb push_back
#define ppb pop_back()
#define fi first
#define se second
#define pi pair<int, int>
#define rep(i, a, n) for(int i = a; i <= n; i++)
#define REP(i, a, n) for(int i = a; i < n; i++)
#define Y "Yes"
#define N "No"
using namespace std;

ll n, k, sum, x[200001], mi, ma;

bool p(ll a) {
	ll cnt = 0;
	REP(i, 0, n) {
		if (x[i] > a) {
			ll y = x[i] - a;
			cnt += (y - 1) / a + 1;
		}
	}
	return cnt <= k;
}

ll search(ll a, ll b) {
	if (b - a <= 1) {
		if (a > b) swap(a, b);
		if (p(a)) return a;
		return a + 1;
	}
	ll m = (a + b) / 2;
	if (p(m)) return search(a, m);
	return search(m, b);
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	//freopen("in.txt", "r", stdin);

	cin >> n >> k >> x[0];
	mi = x[0];
	ma = x[0];
	REP(i, 1, n) {
		cin >> x[i];
		mi = min(mi, x[i]);
		ma = max(ma, x[i]);
	}
	if (k == 0) cout << ma;
	else cout << search(mi / k, ma);
}