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 java.util.*;

import static java.lang.Math.*;

public class Main {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");
    PrintWriter pw = new PrintWriter(System.out);

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

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

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

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

    public Main() throws FileNotFoundException {
    }

    public static void main(String[] args) throws IOException {
        new Main().run();
    }

    long mod = (long) 1e9 + 7;
    int ret;

    int gettwo(long x) {
        int count = 0;
        ArrayList<Long> al = new ArrayList<>();
        while (x > 0) {
            al.add(x % 2);
            if (x % 2 == 1) ret++;
            x /= 2;
        }
        for (int i = 0; i < al.size(); i++) {
            count += al.get(i) * (i + 1);
        }
        return count;
    }

    void run() throws IOException {
        long n = nextLong();
        long p = nextLong();
        int col = 1;
        int w = 100000;
        while (w >= 0) {
            long x = n - p * col;
            if (x <= 0) {
                pw.print(-1);
                pw.close();
                return;
            }
            ret = 0;
            int y = gettwo(x);
            if (ret <= col && y >= col) {
                pw.print(col);
                pw.close();
                return;
            }
            col++;
            w--;
        }
        pw.println(-1);
    }

    class Pair {
        int x, y;

        public Pair(int a, int b) {
            x = a;
            y = b;
        }
    }

    class PairComp implements Comparator<Pair> {

        @Override
        public int compare(Pair o1, Pair o2) {
            if (o1.x == o2.x) return Integer.compare(o1.y, o2.y);
            return Integer.compare(o1.x, o2.x);
        }
    }
}