p-binary

题意翻译

### 题意简述 给定整数 $n,p$，求最小的 $x$ 使得其满足 $$\sum_{1}^{x} (2^k+p)=n$$ 其中 $k$ 可以是任意自然数。 ### 输入格式 一行，$n,p(1\leq n \leq 10^9,-1000\leq p \leq 1000)$。 ### 输出格式 如果找不到这样的 $x$ 请输出 $-1$。 否则输出 $x$。

题目描述

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).

输入输出格式

输入格式

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

输出格式

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.

输入输出样例

输入样例 #1

24 0

输出样例 #1

2

输入样例 #2

24 1

输出样例 #2

3

输入样例 #3

24 -1

输出样例 #3

4

输入样例 #4

4 -7

输出样例 #4

2

输入样例 #5

1 1

输出样例 #5

-1

说明

$ 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.