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.BufferedWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;

public class Test {

  static PrintWriter writer =
      new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

  static int readInt() {
    int ans = 0;
    boolean neg = false;
    try {
      boolean start = false;
      for (int c = 0; (c = System.in.read()) != -1; ) {
        if (c == '-') {
          start = true;
          neg = true;
          continue;
        } else if (c >= '0' && c <= '9') {
          start = true;
          ans = ans * 10 + c - '0';
        } else if (start) break;
      }
    } catch (IOException e) {
    }
    return neg ? -ans : ans;
  }

  static long readLong() {
    long ans = 0;
    boolean neg = false;
    try {
      boolean start = false;
      for (int c = 0; (c = System.in.read()) != -1; ) {
        if (c == '-') {
          start = true;
          neg = true;
          continue;
        } else if (c >= '0' && c <= '9') {
          start = true;
          ans = ans * 10 + c - '0';
        } else if (start) break;
      }
    } catch (IOException e) {
    }
    return neg ? -ans : ans;
  }

  static String readString() {
    StringBuilder b = new StringBuilder();
    try {
      boolean start = false;
      for (int c = 0; (c = System.in.read()) != -1; ) {
        if (Character.isLetterOrDigit(c)) {
          start = true;
          b.append((char) c);
        } else if (start) break;
      }
    } catch (IOException e) {
    }
    return b.toString();
  }

  static int readChars(char[] a, int off) {
    int cnt = 0;
    try {
      boolean start = false;
      for (int c = 0; (c = System.in.read()) != -1; ) {
        if (Character.isLetterOrDigit(c)) {
          start = true;
          a[off + cnt++] = (char) c;
        } else if (start) break;
      }
    } catch (IOException e) {
    }
    return cnt;
  }

  public static void main(String[] args) {
    Test te = new Test();
    te.start();
    writer.flush();
  }

  void start() {
    int n = readInt(), p = readInt();
    for (int i = 1; i < 64; i++) {
      long v = n - p*(long)i;
      if (Long.bitCount(v) <= i && v >= i) {
        writer.println(i);
        return;
      }
    }
    writer.println(-1);
  }
}
