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 PyPy3 (7.3.0) (judge verdict: Accepted)

from bisect import bisect_right
from math import ceil
N,K=map(int, input().split())
A=list(map(int, input().split()))
if K==0:
    print(max(A))
    exit()

A=sorted(A)
# 最小値で二分探索 してまだ行けるかどうか
# 回数で二分探索
left=0
right=max(A)

ans=float("INF")
pre=ans
flag=False
count=0
while True:
    now=0
    mid=(left+right)/2
    idx=bisect_right(A, mid)
    for i in range(idx, N):
        now+=ceil(A[i]/mid)-1
    if now==K:
        if ans!=float("INF"):
            if ans==ceil(mid):
                flag=True
        ans=min(ans, ceil(mid))
        right=mid
        
        #if flag:
        #    break
        pre=ans
    if now>K:
        left=mid
        pre=mid
    if now<K:
        ans=min(ans, ceil(mid))
        right=mid
        pre=mid
    count+=1
    if count>=10**2:
        break
    


if ans==float("INF"):
    ans=max(A)
print(ceil(ans))