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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

def examA():
    X = I()
    if X<30:
        ans = "No"
    else:
        ans = "Yes"
    print(ans)
    return

def examB():
    N, D = LI()
    cnt = 0
    for _ in range(N):
        x, y = LI()
        if x**2+y**2<=D**2:
            cnt += 1
    ans = cnt
    print(ans)
    return

def examC():
    N = 10**7
    K = I()
    cur = 0
    now = 1
    ans = -1
    for i in range(N):
        cur += 7 * now
        cur %= K
        if cur==0:
            ans = i+1
            break
        now *= 10
        now %= K
    print(ans)
    return

def examD():
    N = I()
    C = SI()
    cnt = 0
    l = 0; r = N-1
    while(l<r):
        while(l<N):
            if C[l]=="W":
                break
            l += 1
        while(0<=r):
            if C[r]=="R":
                break
            r -= 1
        if l<r:
            cnt += 1
            l += 1
            r -= 1
    ans = cnt
    print(ans)
    return

def examE():
    N, K = LI()
    A = LI()
    l = 0; r = inf
    for _ in range(150):
        now = dec((l+r)/2)
        cnt = 0
        for a in A:
            cur = a//now
            cnt += cur
        if K<cnt:
            l = now
        else:
            r = now
        #print(now,cnt)
        #input()
    ans = now
    #print(now)
    if now%1>0:
        ans = (ans+1)//1
    else:
        ans = ans//1
    print(ans)
    return

def examF():
    N, Q = LI()
    C = LI()
    for q in range(Q):
        l, r = LI()


    ans = [0]*Q
    for v in ans:
        print(v)
    return

from decimal import getcontext,Decimal as dec
import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep,alphabet_convert
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = dec("0.000000000001")
alphabet = [chr(ord('a') + i) for i in range(26)]
alphabet_convert = {chr(ord('a') + i): i for i in range(26)}

getcontext().prec = 28

sys.setrecursionlimit(10**7)

if __name__ == '__main__':
    examE()

"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
"""