CF1225C.p-binary

Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer $p$ (which may be positive, negative, or zero). To combine their tastes, they invented $p$-binary numbers of the form $2^x + p$, where $x$ is a non-negative integer.

For example, some $-9$-binary ("minus nine" binary) numbers are: $-8$ (minus eight), $7$ and $1015$ ($-8=2^0-9$, $7=2^4-9$, $1015=2^{10}-9$).

The boys now use $p$-binary numbers to represent everything. They now face a problem: given a positive integer $n$, what's the smallest number of $p$-binary numbers (not necessarily distinct) they need to represent $n$ as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.

For example, if $p=0$ we can represent $7$ as $2^0 + 2^1 + 2^2$.

And if $p=-9$ we can represent $7$ as one number $(2^4-9)$.

Note that negative $p$-binary numbers are allowed to be in the sum (see the Notes section for an example).

瓦西娅喜欢任意一个形如 $2^x$（其中 $x$ 为非负整数）的整数，即任意一个 2 的非负整数次幂。而佩佳则非常保守，只喜欢某一个整数 $p$（$p$ 可以是正数、负数或零）。为了融合彼此的喜好，他们定义了所谓的 $p$-进制数（$p$-binary numbers），其形式为 $2^x + p$，其中 $x$ 是一个非负整数。

例如，一些 $-9$-进制数（“负九”进制数）包括：$-8$（负八）、$7$ 和 $1015$（因为 $-8 = 2^0 - 9$，$7 = 2^4 - 9$，$1015 = 2^{10} - 9$）。

现在，这两个男孩用 $p$-进制数来表示所有事物。他们面临一个问题：给定一个正整数 $n$，最少需要多少个 $p$-进制数（允许重复使用）相加才能得到 $n$？也有可能根本无法表示 $n$。请帮助他们解决这个问题。

例如，若 $p = 0$，我们可以将 $7$ 表示为 $2^0 + 2^1 + 2^2$。

又如，若 $p = -9$，我们可以将 $7$ 表示为单个 $p$-进制数 $(2^4 - 9)$。

注意：求和中允许出现负的 $p$-进制数（参见“说明”部分中的一个例子）。

The only line contains two integers $n$ and $p$ ($1 \leq n \leq 10^9$, $-1000 \leq p \leq 1000$).

唯一一行包含两个整数 $n$ 和 $p$（$1 \leq n \leq 10^9$，$-1000 \leq p \leq 1000$）。

If it is impossible to represent $n$ as the sum of any number of $p$-binary numbers, print a single integer $-1$. Otherwise, print the smallest possible number of summands.

如果无法将 $n$ 表示为任意个 $p$-二进制数的和，则输出单个整数 $-1$；否则，输出可能的最小加数个数。

$0$-binary numbers are just regular binary powers, thus in the first sample case we can represent $24 = (2^4 + 0) + (2^3 + 0)$.

In the second sample case, we can represent $24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1)$.

In the third sample case, we can represent $24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1)$. Note that repeated summands are allowed.

In the fourth sample case, we can represent $4 = (2^4 - 7) + (2^1 - 7)$. Note that the second summand is negative, which is allowed.

In the fifth sample case, no representation is possible.

$0$-进制数即为普通的二进制幂，因此在第一个样例中，我们可以表示 $24 = (2^4 + 0) + (2^3 + 0)$。

在第二个样例中，我们可以表示 $24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1)$。

在第三个样例中，我们可以表示 $24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1)$。注意：允许重复的加数。

在第四个样例中，我们可以表示 $4 = (2^4 - 7) + (2^1 - 7)$。注意：第二个加数为负数，这是允许的。

在第五个样例中，不存在可行的表示方式。