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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

int f(const void *b, const void *a)
{
    return (*(int *)a - *(int *)b);
}



int main(void){
    // Your code here!
    long long int n, k, a[200001], tmp, c[200001] = {},i, ii;
    scanf("%lld%lld", &n, &k);
    for (i = 0; i < n; i++)
        scanf("%lld", a + i);
    qsort(a, n, sizeof(long long int), f);
    for (i = 0; i < k; i++)
    {
        c[0]++;
        for (ii = 0; ii < n - 1; ii++)
        {
            if (a[ii] *(c[ii + 1] +1) < a[ii + 1] * (c[ii] + 1))
            {
                tmp = a[ii];
                a[ii] = a[ii + 1];
                a[ii + 1] = tmp;
                tmp = c[ii];
                c[ii] = c[ii + 1];
                c[ii + 1] = tmp;
                
            }
            else
            break;
        }
    }
    printf("%lld", a[0] / (c[0] + 1) + (a[0] % (c[0] + 1) > 0));
    

}