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: Accepted)

from heapq import heapify,heappop,heappush
from math import ceil

def solve(A,N,K):
    ans_max,ans_min = max(A),1

    if cutable(A,N,K,1):
        return 1

    while True:
        if cutable(A,N,K,(ans_max+ans_min)//2):
            if ans_max > (ans_max+ans_min)//2:
                ans_max = (ans_max+ans_min)//2
            else:
                return ans_max
        else:
            if ans_min < (ans_max+ans_min)//2:
                ans_min = (ans_max+ans_min)//2
            else:
                return ans_max

def cutable(A,N,K,div):
    #K回までの切断で全部を長さdiv以下にできるかどうか
    cnt = 0
    for i in range(N):
        cnt += A[i]//div
        if A[i] % div == 0:
            cnt -= 1
        if cnt > K:
            return False

    return True

def main():
    N,K = map(int,input().split())
    A = list(map(int,input().split()))

    ans = solve(A,N,K)
    print(ans)

if __name__ == "__main__":
    main()