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: Wrong Answer)

#各丸太を何回切るか、どこで切るか
#丸太iをKi回切る場合、その丸太の最小はL//Kiとなる
#あとは、どの丸太を何回切るかを求めればよい。
"""
長い丸太ほど多く切るべき
丸太が2本、1本が10**9でもう一本が1。10**9回切れるのであれば1本目に全振りするのがベスト。

結論、元の長さの比に基づいて割り当てればよい。
一番短い丸太の長さを1として各丸太の
"""
from math import ceil
import heapq
N,K = map(int,input().split())
A = list(map(int,input().split()))
A.sort()
pivot = A[0]
hiritu = [int(ceil(a/pivot)) for a in A]
hirituSum = sum(hiritu)
kaisu = []
for i in range(N):
    h = hiritu[i]
    k = int(K*h/hirituSum)
    kaisu.append(k)
    K -= k

if K >= 1000000:
    for i in range(N):
        h = hiritu[i]
        k = int(ceil(K*h/hirituSum))
        kaisu.append(k)
        K -= k

leng = []
for i in range(N):
    a = A[i]
    k = kaisu[i]
    leng.append(((-a/(k+1)),k,a))
heapq.heapify(leng)
for _ in range(K):
    l,k,a = heapq.heappop(leng)
    l = -a/(k+2)
    heapq.heappush(leng,(l,k+1,a))
l,k,a = heapq.heappop(leng)
l *= -1
print(ceil(l))