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)

import math
N, K = [int(_) for _ in input().split()]
A = sorted([int(_) for _ in input().split()])

#x回切ると1/(x+1)倍になる


def check(x):
    if x == 0:
        print(1)
        exit()
    cnt = sum(0 - (-a) // x for a in A) - N
    return cnt <= K


lb = -1
rb = 10**9 + 1
# Max x for check(x) == True
while rb - lb > 1:
    mid = (rb + lb) // 2
    if check(mid):
        rb = mid
    else:
        lb = mid
ans = lb
d = 0
for a in A:
    if a % ans:
        d = 1
        break
ans += d
print(ans)