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.*;

public class Main {


    public static void main(String[] args) throws IOException {
//        br = new BufferedReader(new FileReader(new File("cutting.in")));
//        out = new PrintWriter(new File("cutting.out"));
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        int PUTEN = 1;
        for (int POMOGI = 0; POMOGI < PUTEN; POMOGI++) {
            int n = nextInt();
            int k = nextInt();
            if(k + 1 > n){
                out.println(-1);
                continue;
            }
            int[] a = new int[31];
            int st = 1;
            int ind = 0;
            while (ind < 31) {
                a[ind] = st;
                ind++;
                st *= 2;
            }
            int otvet = -1;
            for (int ans = 1; ans < 10000; ans++) {
                int popka = n - (k * ans);
                if(popka < ans)continue;
                int cnt = 0;
                ind = 30;
                while (ind >= 0 && popka != 0) {
                    if (popka - a[ind] < 0) ind--;
                    else {
                        popka -= a[ind];
                        cnt++;
                    }
                }
                if (ind != -1 && cnt <= ans && otvet == -1) otvet = ans;
            }
            out.println(otvet);
        }
        out.close();
    }


    static BufferedReader br;
    static PrintWriter out;
    static StringTokenizer in = new StringTokenizer("");

    static String next() throws IOException {
        while (!in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

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

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

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

    static int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }
}