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 org.omg.PortableInterceptor.INACTIVE;

import java.io.*;
import java.util.*;

public class Solver {
    static long gcd(long a, long b) {
        return a == 0 ? b : gcd(b, b % a);
    }

    static long binpow(long a, int n) {
        return n == 1 ? a : (n & 1) == 0 ? binpow(a, n / 2) * binpow(a, n / 2) : a * binpow(a, n - 1);
    }

    static long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }
    static int p;
    static long a[];
    static int calcMin(long n){
        int cnt = 0;
        while (n != 1){
            if(n % 2 == 0)n/=2;
            else {
                n--;
                cnt++;
            }
        }
        return cnt + 1;
    }
    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        // FastScanner in = new FastScanner("search.in");
        //PrintWriter out = new PrintWriter("search.out");
        long n = in.nextLong();
        p = in.nextInt();
        int ans = -1;
        for (int i = 1; i <= 100000; i++) {
          if(n - p * i <= 0)break;
          if(calcMin(n - p*i) <= i && n - p * i >= i){
              ans = i ;
              break;
          }
        }
        out.println(ans);
        out.close();
    }
}
class FastScanner {
    BufferedReader br;
    StringTokenizer st = new StringTokenizer("");

    public FastScanner(InputStream s) {
        br = new BufferedReader(new InputStreamReader(s));
    }

    public FastScanner(String s) throws FileNotFoundException {
        br = new BufferedReader(new FileReader(new File(s)));
    }

    public String next() throws IOException {
        if (!st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
}