#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 pp (which may be positive, negative, or zero). To combine their tastes, they invented pp-binary numbers of the form 2x+p2^x + p, where xx is a non-negative integer.

For example, some 9-9-binary ("minus nine" binary) numbers are: 8-8 (minus eight), 77 and 10151015 (8=209-8=2^0-9, 7=2497=2^4-9, 1015=21091015=2^{10}-9).

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

For example, if p=0p=0 we can represent 77 as 20+21+222^0 + 2^1 + 2^2.

And if p=9p=-9 we can represent 77 as one number (249)(2^4-9).

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

The only line contains two integers nn and pp (1n1091 \leq n \leq 10^9, 1000p1000-1000 \leq p \leq 1000).

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

Input

The only line contains two integers nn and pp (1n1091 \leq n \leq 10^9, 1000p1000-1000 \leq p \leq 1000).

Output

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

Samples

Sample Input 1

24 0

Sample Output 1

2

Sample Input 2

24 1

Sample Output 2

3

Sample Input 3

24 -1

Sample Output 3

4

Sample Input 4

4 -7

Sample Output 4

2

Sample Input 5

1 1

Sample Output 5

-1

Note

00-binary numbers are just regular binary powers, thus in the first sample case we can represent 24=(24+0)+(23+0)24 = (2^4 + 0) + (2^3 + 0).

In the second sample case, we can represent 24=(24+1)+(22+1)+(20+1)24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).

In the third sample case, we can represent 24=(241)+(221)+(221)+(221)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=(247)+(217)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.