CF1213D1.Equalizing by Division (easy version)

The only difference between easy and hard versions is the number of elements in the array.

You are given an array $a$ consisting of $n$ integers. In one move you can choose any $a_i$ and divide it by $2$ rounding down (in other words, in one move you can set $a_i := \lfloor\frac{a_i}{2}\rfloor$).

You can perform such an operation any (possibly, zero) number of times with any $a_i$.

Your task is to calculate the minimum possible number of operations required to obtain at least $k$ equal numbers in the array.

Don't forget that it is possible to have $a_i = 0$ after some operations, thus the answer always exists.

简单版本与困难版本的唯一区别在于数组中元素的个数。

给定一个由 $n$ 个整数组成的数组 $a$。在一次操作中，你可以任选一个元素 $a_i$ 并将其除以 $2$ 后向下取整（即，一次操作可将 $a_i$ 更新为 $a_i := \lfloor\frac{a_i}{2}\rfloor$）。

你可以对任意 $a_i$ 执行上述操作任意次（包括零次）。

你的任务是计算：为使数组中至少出现 $k$ 个相等的数，所需执行的最少操作次数。

请注意，经过若干次操作后可能出现 $a_i = 0$，因此答案恒存在。

The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 50$) — the number of elements in the array and the number of equal numbers required.

The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$), where $a_i$ is the $i$-th element of $a$.

输入的第一行包含两个整数 $n$ 和 $k$（$1 \le k \le n \le 50$）——分别表示数组中元素的个数以及所需相等数字的个数。

输入的第二行包含 $n$ 个整数 $a_1, a_2, \dots, a_n$（$1 \le a_i \le 2 \cdot 10^5$），其中 $a_i$ 是数组 $a$ 的第 $i$ 个元素。

Print one integer — the minimum possible number of operations required to obtain at least $k$ equal numbers in the array.

输出一个整数——在数组中获得至少 $k$ 个相等数字所需的最少操作次数。