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.
#include <bits/stdc++.h>
using namespace std;
long long pot[35];
int numbits(long long n) {
  int conta = 0;
  for (int i = 0; i <= 33; i++) {
    if (n & pot[i]) conta++;
  }
  return conta;
}
int main() {
  pot[0] = 1;
  for (int i = 1; i <= 33; i++) pot[i] = 2 * pot[i - 1];
  long long n, p;
  scanf("%lld %lld", &n, &p);
  if (p == 0)
    printf("%d\n", numbits(n));
  else if (p < 0) {
    int k = 0;
    bool teste = true;
    while (teste) {
      if (numbits(n - k * p) <= k) {
        printf("%d\n", k);
        teste = false;
      } else
        k++;
    }
  } else {
    int k = 0;
    bool funfa = false;
    while (k <= (n / (p + 1)) && funfa == false) {
      if (numbits(n - k * p) <= k) {
        printf("%d\n", k);
        funfa = true;
      }
      k++;
    }
    if (!funfa) printf("-1\n");
  }
}
