#P1225C. p-binary

p-binary

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.