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.Scanner;
import java.util.StringTokenizer;


public class Main {

    FastScanner in;
    PrintWriter out;

    int dif(int a, int b){
        if(a == 9 && b == 1) return 1;
        return b - a;
    }

    private void solve() throws IOException{
        int n = in.nextInt(), p = in.nextInt();
        if(p == 0){
            String ch = Integer.toBinaryString(n);
            int am = 0;
            for(int j = 0; j < ch.length(); j++){
                if(ch.charAt(j) == '1'){
                    am++;
                }
            }
            out.print(am);
            return;
        }
        int cnt = 0;
        for(int i = 0; i < 50 && n > p; i++){
            cnt++;
            n -= p;
            String ch = Integer.toBinaryString(n);
            int am = 0;
            for(int j = 0; j < ch.length(); j++){
                if(ch.charAt(j) == '1'){
                    am++;
                }
            }
            if(am <= cnt && n >= cnt){
                out.print(cnt);
                return;
            }

        }
        out.print(-1);
    }

    private void run() throws IOException{
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);

        solve();
        out.close();
    }

    class FastScanner {
        StringTokenizer st;
        BufferedReader br;
        FastScanner(InputStream s){
            br = new BufferedReader(new InputStreamReader(s));
        }

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

        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
        long nextLong() throws IOException {
            return Long.parseLong(next());
        }
    }

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

	// write your code here
    }
}