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).

Input

The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 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.

Examples

Input


24 0


Output


2


Input


24 1


Output


3


Input


24 -1


Output


4


Input


4 -7


Output


2


Input


1 1


Output


-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.
import java.io.*;
import static java.lang.Double.*;
import static java.lang.Math.pow;
import java.util.*;
public class MoreMore {
    static FastReader in =new FastReader();
    public static void main(String[] args) throws IOException {
        long n=in.nextLong(),p=in.nextLong(),sum=0,lol=0;
        if(n<=p){System.out.println("-1");return;}
        while(sum<=n){
            long lop=n-sum,k=0,look=0,kk=0;
            while(lop>0){
            if(lop%2==1){k++;kk+=pow(2,look);}
            lop/=2;
           
            look++;
            }
            if(k<=lol&&lol<=kk){System.out.println(lol);return;}
            
            lol++;sum+=p;
            
        }
        System.out.println("-1");
    }
    static long gcd(long g,long x){
        if(x<1)return g;
        else return gcd(x,g%x);
    }
}
class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(new
                     InputStreamReader(System.in));
        }
 
        String next() throws IOException
        {
            while (st == null || !st.hasMoreElements())
            {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }
 
        int nextInt() throws IOException
        {
            return Integer.parseInt(next());
        }
 
        long nextLong() throws IOException
        {
            return Long.parseLong(next());
        }
 
        double nextDouble() throws IOException
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
}