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 <stdio.h>
#define max(a, b) ((a) > (b) ? (a) : (b))

int n, k;
int a[200010];
char check(const int n);

int main() {
	scanf("%d%d", &n, &k);
	int maxv = 0;
	long long sum = 0;
	for (int i = 0; i < n; ++i) {
		scanf("%d", &a[i]);
		sum += a[i], maxv = max(maxv, a[i]);
	}
	
	int l = (sum + n + k - 1) / (n + k), r = maxv;
	while (l < r) {
		int m = l + r >> 1;
		if (check(m))
			r = m;
		else
			l = m + 1;
	}
	printf("%d\n", l);
	return 0;
}

char check(const int m) {
	int sum = 0;
	for (int i = 0; i < n; ++i)
		sum += 2 * a[i] / (2 * m + 1) + 1;
	return sum <= n + k;
}