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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.io.InputStream;

/**
 * @author khokharnikunj8
 */

public class Main {
    public static void main(String[] args) {
        new Thread(null, new Runnable() {
            public void run() {
                new Main().solve();
            }
        }, "1", 1 << 26).start();
    }

    void solve() {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        ScanReader in = new ScanReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        APBinary solver = new APBinary();
        solver.solve(1, in, out);
        out.close();
    }

    static class APBinary {
        public void solve(int testNumber, ScanReader in, PrintWriter out) {
            long n = in.scanInt();
            long p = in.scanInt();
            for (long i = 1; i <= 100000000; i++) {
                n -= i * p;
                if (n < 0) {
                    n += i * p;
                    continue;
                }
                if (n >= i && Long.bitCount(n) <= i) {
                    out.println(i);
                    return;
                }
                n += i * p;
            }
            out.println(-1);
        }

    }

    static class ScanReader {
        private byte[] buf = new byte[4 * 1024];
        private int index;
        private BufferedInputStream in;
        private int total;

        public ScanReader(InputStream inputStream) {
            in = new BufferedInputStream(inputStream);
        }

        private int scan() {
            if (index >= total) {
                index = 0;
                try {
                    total = in.read(buf);
                } catch (Exception e) {
                    e.printStackTrace();
                }
                if (total <= 0) return -1;
            }
            return buf[index++];
        }

        public int scanInt() {
            int integer = 0;
            int n = scan();
            while (isWhiteSpace(n)) n = scan();
            int neg = 1;
            if (n == '-') {
                neg = -1;
                n = scan();
            }
            while (!isWhiteSpace(n)) {
                if (n >= '0' && n <= '9') {
                    integer *= 10;
                    integer += n - '0';
                    n = scan();
                }
            }
            return neg * integer;
        }

        private boolean isWhiteSpace(int n) {
            if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true;
            else return false;
        }

    }
}

