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 <stdlib.h>
#include <string.h>
#include <math.h>
#include <stdbool.h>
typedef long long ll;
#define rep(i,n) for(ll i=0; i<(ll)(n); i++)
#define max(l,r) ((l)>(r)?l:r)
#define min(l,r) ((l)<(r)?l:r)
#define swap(l,r) {ll tp=l;l=r;r=tp;}
const ll MOD = 1000000007;
int upcmpii(const void *a, const void *b){
    return (*(int*)a == *(int*)b ? 0 : (*(int*)a > *(int*)b ? 1 : -1));
}
void sort_ui(int *a, int n){ qsort(a, n, sizeof(int), upcmpii);}

int main(){
    //input
    int n, k;
    scanf("%d%d", &n, &k);
    int a[201010];
    rep(i, n) scanf("%d", a+i);

    //solve
    sort_ui(a, n);
    if(k == 0) {
        printf("%d\n", a[n-1]);
    }else{
        int left = 0, right = a[n-1]+1;
        while(right - left > 1){
            ll tmp = 0;
            int mid = (right+left)/2;
            rep(i, n){
                if(a[i] > mid) {
                    tmp += (a[i]-1)/mid;
                }
            }
            if(tmp > k) left = mid;
            else right = mid;
        }
        printf("%d", right);
    }
}