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: Wrong Answer)

import sys
sys.setrecursionlimit(10 ** 9)
# input = sys.stdin.readline    ####
def int1(x): return int(x) - 1
def II(): return int(input())
def MI(): return map(int, input().split())
def MI1(): return map(int1, input().split())
def LI(): return list(map(int, input().split()))
def LI1(): return list(map(int1, input().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def MS(): return input().split()
def LS(): return list(input())
def LLS(rows_number): return [LS() for _ in range(rows_number)]
def printlist(lst, k=' '): print(k.join(list(map(str, lst))))
INF = float('inf')
from math import ceil, floor, log2
# from collections import deque
# from itertools import combinations as comb, combinations_with_replacement as comb_w, accumulate, product, permutations
# from heapq import heapify, heappop, heappush
# import numpy as np    # cumsum
# from bisect import bisect_left, bisect_right

def solve():
    N, K = MI()
    A = LI()
    A.sort()
    # print(A)
    
    l = 0
    r = max(A)+1
    while l != r-1:
        x = (l + r) // 2
        cnt = 0
        for a in A:
            cnt += ceil(a / x) - 1
        if cnt > K:
            l = x
        else:
            r = x
        # print(x, cnt, l, r)
    print(x)

if __name__ == '__main__':
    solve()