#P1225C. p-binary

    ID: 2168 Type: RemoteJudge 2000ms 512MiB Tried: 0 Accepted: 0 Difficulty: (None) Uploaded By: Tags>bitmasksbrute forcemath*1600

p-binary

No submission language available for this problem.

Description

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).

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

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.

Input

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

Output

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.

Samples

24 0
2
24 1
3
24 -1
4
4 -7
2
1 1
-1

Note

$0$-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.