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.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class A {

	static long N,P;
	public static void main(String[] args) {
		FS in = new FS();
		N = in.nextLong();
		P = in.nextLong();
		long forBit[] = new long[70];
		forBit[0] = 1;
		for(int i = 1; i < 70; i++) {
			forBit[i] = 2*forBit[i-1];
		}
		
		
		for(int terms = 1; terms <= 100; terms++) {
			long goal = N-(P*terms);
			if(goal <= 0) continue;
			long min = Long.bitCount(goal);
			long max = 0;
			for(long i = 0; i < 63; i++) {
				if((goal & (1L<<i)) > 0) {
					max += (forBit[(int)i]);
				}
			}
//			System.out.println(terms+" "+goal+" "+min+" "+max);
			if(terms >= min && terms <= max) {
				System.out.println(terms);
				return;
			}
		}
		System.out.println(-1);
	}
	static class FS{
		BufferedReader br;
		StringTokenizer st;
		public FS(String s) throws Exception {
			try{ br = new BufferedReader(new FileReader(s)); }
			catch(Exception e) { throw new Exception(); }
		}
		public FS(){
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String nextToken()	{
			while(st == null ||!st.hasMoreElements()){
				try { st = new StringTokenizer(br.readLine());}
				catch(IOException e) { e.printStackTrace(); }			
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(nextToken());
		}
		long nextLong() {
			return Long.parseLong(nextToken());
		}
		double nextDouble() {
			return Double.parseDouble(nextToken());
		}
		String next() {
			return nextToken();
		}

	}
	
}
