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 Python (3.8.2) (judge verdict: Runtime Error)

nk=input("").split(" ")
n=int(nk[0])
k=int(nk[1])
lia=input("").split(" ")
lisa=[]
for i in lia:
    lisa+=[int(i)*10000]
sa=sum(lisa)
lisf=[]
lisw=[]
t=int(sa/(k+1))

for i in lisa:
    f=i%t
    ffff=(i*(k+1))%sa
    
    if(ffff==0):
        lisf+=[0]
        lisw+=[int(i*(k+1)/sa)-1]
    else:
        lisf+=[(sa/(k+1)-f)/(int(i*(k+1)/sa)+1)]
        lisw+=[int(i*(k+1)/sa)]

lisf.sort()
s=k-sum(lisw)

kk=int((t-lisf[s])/10000)
if(((sa/(k+1)-lisf[s])%10000)==0):
    print(kk)
else:
    print(kk+1)