Amr loves Chemistry, and specially doing experiments. He is preparing for a new interesting experiment.

Amr has *n* different types of chemicals. Each chemical *i* has an initial volume of *a*<sub class="lower-index">*i*</sub> liters. For this experiment, Amr has to mix all the chemicals together, but all the chemicals volumes must be equal first. So his task is to make all the chemicals volumes equal.

To do this, Amr can do two different kind of operations. 
 -  Choose some chemical *i* and double its current volume so the new volume will be 2*a*<sub class="lower-index">*i*</sub> -  Choose some chemical *i* and divide its volume by two (integer division) so the new volume will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/527470887c95577094d0c6c100f6bec4842e6d29.png" style="max-width: 100.0%;max-height: 100.0%;"/> 
Suppose that each chemical is contained in a vessel of infinite volume. Now Amr wonders what is the minimum number of operations required to make all the chemicals volumes equal?
The first line contains one number *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>), the number of chemicals.

The second line contains *n* space separated integers *a*<sub class="lower-index">*i*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>), representing the initial volume of the *i*-th chemical in liters.
import os, sys
from io import BytesIO, IOBase
from types import GeneratorType

def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

class TailRecurseException(BaseException):
    def __init__(self, args, kwargs):
        self.args = args
        self.kwargs = kwargs

def tail_call_optimized(g):
    """
    This function decorates a function with tail call
    optimization. It does this by throwing an exception
    if it is it's own grandparent, and catching such
    exceptions to fake the tail call optimization.
    This function fails if the decorated
    function recurses in a non-tail context.
    """
    def func(*args, **kwargs):
        f = sys._getframe()
        if f.f_back and f.f_back.f_back \
            and f.f_back.f_back.f_code == f.f_code:
            # 抛出异常
            raise TailRecurseException(args, kwargs)
        else:
            while 1:
                try:
                    return g(*args, **kwargs)
                except TailRecurseException as e:
                    args = e.args
                    kwargs = e.kwargs
    func.__doc__ = g.__doc__
    return func

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, 8192))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
if sys.hexversion == 50858480:
    sys.stdin = open('/Users/zhang-wangz/Desktop/algorithm/tea_0x3f/00_cfinput.txt')

def queryInteractive(a, b, c):
    print('? {} {} {}'.format(a, b, c))
    sys.stdout.flush()
    return int(input())

def answerInteractive(x1, x2):
    print('! {} {}'.format(x1, x2))
    sys.stdout.flush()

input = lambda: sys.stdin.readline().strip()
ints = lambda: list(map(int, input().split()))
Int = lambda: int(input())
inf = float('inf')

# https://codeforces.com/problemset/problem/558/C

# 输入 n(≤1e5) 和一个长为 n 的数组 a (1≤a[i]≤1e5)。
# 每次操作你可以将某个 a[i] 乘二或除二（下取整）。
# 输出至少需要多少次操作，可以让 a 的所有数都相同。

def solve():
    n = Int()
    a = ints()
    m = max(a) + 1
    # 记录x已经存在多少个了
    cnt = [0] * m
    # 所有数变成x花了多少步
    steps = [0] * m
    # 是否和上次x时的i一致，作为退出条件
    vis = [-1] * m
    # 遍历
    for i, v in enumerate(a):
        q = [(v, 0)]
        step = 0
        while step < len(q):
            v, s = q[step]
            step += 1
            if v >= m or vis[v] == i:
                continue
            vis[v] = i
            steps[v] += s
            cnt[v] += 1
            q.append((v * 2, s + 1))
            q.append((v // 2, s + 1))
    ans = min(steps[x] for x in range(m) if cnt[x] == n)
    print(ans)

solve()