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 <iostream>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <string>
#include <sstream>
#include <algorithm>
#include <random>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>
#include <cassert>
#include <climits>
#include <bitset>
#include <functional>
#include <iomanip>
#include <random>

#define FOR_LT(i, beg, end) for (int i = (int)(beg); i < (int)(end); i++)
#define FOR_LE(i, beg, end) for (int i = (int)(beg); i <= (int)(end); i++)
#define FOR_DW(i, beg, end) for (int i = (int)(beg); (int)(end) <= i; i--)
#define REP(n)              for (int repeat_index = 0; repeat_index < (int)n; repeat_index++)

using namespace std;

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout << fixed << setprecision(20);

	int n, k; cin >> n >> k;
	vector<int> as(n);
	int maxv = 0;
	FOR_LT(i, 0, n) {
		cin >> as[i];
		maxv = max(maxv, as[i]);
	}

	int64_t ng = 1;
	int64_t ok = maxv;
	while (ng + 1 < ok) {
		int64_t mid = (ng + ok) / 2;
		int64_t ck = 0;
		FOR_LT(i, 0, n) {
			ck += (as[i] + mid - 1) / mid - 1;
		}

		if (ck <= k) {
			ok = mid;
		}
		else {
			ng = mid;
		}
	}
	cout << ok << endl;
	

	return 0;
}