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: Accepted)

#include <stdio.h>
#include <math.h>
typedef long long LL;
typedef long double Lf;
#define in (void)scanf
#define rep(i,n)for(int i=0;i<(int)(n);i++)
#define MOD 1000000007
int n, k, a[200009];
LL check(int x) {
	LL ret = 0;
	rep(i, n)ret += (LL)ceill((Lf)a[i] / (Lf)x) - 1;
	return ret;
}

int main(void) {
	in("%d%d", &n, &k);
	rep(i, n)in("%d", &a[i]);
	int p = 0, q = MOD / 2, r = MOD, ans = n;
	while (r - p > 1) {
		if (check(q) <= (LL)k) r = q;
		else p = q;
		q = (p + r) / 2;
	}
	if (check(r) <= k)ans = r;
	if (check(q) <= k)ans = q;
	if (check(p) <= k)ans = p;
	printf("%d\n", ans ? ans : 1);
	return 0;
}