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)

#coding: utf-8
N, K = (int(n) for n in input().split())
A = [int(n) for n in input().split()]

def is_ok(X):
    ret = 0
    for log in A:
        ret += -(-log//X) - 1
    return ret <= K

def meguru_bisect(ng, ok):
    '''
    初期値のng,okを受け取り,is_okを満たす最小(最大)のokを返す
    まずis_okを定義すべし
    ng ok は  とり得る最小の値-1 とり得る最大の値+1
    最大最小が逆の場合はよしなにひっくり返す
    '''
    while (abs(ok - ng) > 1):
        mid = (ok + ng) // 2
        if is_ok(mid):
            ok = mid
        else:
            ng = mid
    return ok

print(meguru_bisect(0, max(A)+1))